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# Arithmetic Progressions : Exercise 5.3 (Mathematics NCERT Class 10th)

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Q.1      Find the sum of the following APs :
(i) 2, 7, 12, .... to 10 terms.
(ii) – 37, –33, – 29, .... to 12 terms.
(iii) 0.6, 1.7, 2.8...., to 100 terms.
(iv) ${1 \over {15}},{1 \over {12}},{1 \over {10}},....$ to 11 terms.
Sol.        (i) Let a be the first term and d be the common difference of the given AP then, we have
a = 2 and d = 7 – 2 = 5
We have to find the sum of 10 terms of the given AP.
Putting a = 2, d = 5, n = 10 in ${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]$
we get
${S_{10}} = {{10} \over 2}\left[ {2 \times 2 + \left( {10 - 1} \right)5} \right]$
$= 5\left( {4 + 9 \times 5} \right)$
= 5(4 + 45) = 5 × 49 = 245

(ii) Let a be the first term and d be the common difference of the given AP. Then we have
a = – 37, d = – 33 – (– 37) = – 33 + 37 = 4
We have to find the sum of 12 terms of the given AP.
Putting a = – 37, d = 4, n = 12 in
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right],we\,\,get$
${S_{12}} = {{12} \over 2}\left[ {2 \times - 37 + 12\left( {12 - 1} \right)4} \right]$
= 6 (– 74 + 11 × 4)
= 6 (– 74 + 44) = 6 × (– 30) = – 180

(iii) Let a be the first term and d be the common difference of the given AP. Then, we have
a = 0.6, d = 1.7 – 0.6 = 1.1
We have to find the sum of 100 terms of the given AP
Putting a = 0.6 d = 1.1, n = 100 in
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right],\,we\,get$
${S_{100}} = {{100} \over 2}\left[ {2 \times 0.6 + \left( {100 - 1} \right)1.1} \right]$
= 50(1.2 + 99 × 1.1)
= 50 (1.2 + 108.9)
= 50 × 110.1 = 5505

(iv) Let a be the first term and d be the common difference of the given AP. Then we have
$a = {1 \over {15}},d = {1 \over {12}} - {1 \over {15}} = {{5 - 4} \over {60}} = {1 \over {60}}$
We have to find the sum of 11 terms of the given AP.
Putting $a = {1 \over {15}},d = {1 \over {60}},n = 11\,in$
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right],\,we\,get$
${S_{11}} = {{11} \over 2}\left[ {2 \times {1 \over {15}} + \left( {11 - 1} \right){1 \over {60}}} \right]$
$= {{11} \over 2}\left( {{2 \over {15}} + 10 \times {1 \over {60}}} \right)$
$= {{11} \over 2}\left( {{2 \over {15}} + {1 \over 6}} \right)$
$= {{11} \over 2} \times {{4 + 5} \over {30}}$
$= {{11} \over 2} \times {9 \over {30}} = {{33} \over {20}}$

Q.2      Find the sums given below :
(i) $7 + 10{1 \over 2} + 14 + .... + 84$
(ii) 34 + 32 + 30 + ...... + 10
(iii) – 5 + (– 8) + (– 11) + .... + (– 230)
Sol.        (i) Here, the last term is given. We will first have to find the number of terms.
$a = 7,\,d = 10{1 \over 2} - 7 = 3{1 \over 2} = {7 \over 2},\ell = {a_n} = 84$
Therefore, 84 = a + (n – 1) d
$\Rightarrow$ $84 = 7 + \left( {n - 1} \right){7 \over 2}$
$\Rightarrow$ ${7 \over 2}\left( {n - 1} \right) = 84 - 7$
$\Rightarrow$ ${7 \over 2}\left( {n - 1} \right) = 77$
$\Rightarrow$ $n - 1 = 77 \times {2 \over 7}$
$\Rightarrow$ n – 1 = 22
$\Rightarrow$ n = 23
We know that
${S_n} = {n \over 2}\left( {a + \ell } \right)$
$\Rightarrow$ ${S_{23}} = {{23} \over 2}\left( {7 + 84} \right) = {{23} \over 2} \times 91$
$= {{2093} \over 2} = 1046{1 \over 2}$

(ii) Here, the last term is given. We will first have to find the number of terms.
a = 34, d = 32 – 34 = – 2, l = ${a_n} = 10$
Therefore 10 = a + (n – 1)d
$\Rightarrow$ 10 = 34 + (n – 1) (– 2)
$\Rightarrow$ (– 2) (n – 1) = 10 – 34
$\Rightarrow$ (– 2) (n –1) = – 24
$\Rightarrow$ n – 1 = 12
$\Rightarrow$ n = 12 + 1 = 13
Using ${S_n} = {n \over 2}\left( {a + \ell } \right)$, we have
${S_{13}} = {{13} \over 2}\left( {34 + 10} \right) = {{13} \over 2} \times 44$ = 13 × 22 = 286

(iii) Here the last term is given. We will first have to find the number of terms.
a = – 5, d = – 8 – (–5) = – 8 + 5 = – 3, l = ${a_n} = - 230$
Therefore – 230 = a + (n – 1) d
$\Rightarrow$ – 230 = – 5 + (n – 1) (– 3)
$\Rightarrow$ (– 3) (n – 1) = – 230 + 5
$\Rightarrow$ (– 3) (n – 1) = – 225
$\Rightarrow$ $n - 1 = {{ - 225} \over { - 3}}$
$\Rightarrow$ n – 1 = 75
$\Rightarrow$ n = 75 + 1 = 76
Using ${S_n} = {n \over 2}\left( {a + \ell } \right)$, we have
${S_{76}} = {{76} \over 2}\left( { - 5 - 230} \right)$ = 38 × – 235 = – 8930

Q.3      In an AP :
(i) Given a = 5, d = 3, ${a_n} = 50,\,find\,n\,and\,{S_n}$.
(ii) Given a = 7, ${a_{13}} = 35,\,find\,d\,and\,{S_{13}}$
(iii)Given ${a_{12}} = 37,\,d = 3,\,find\,a\,and\,{S_{12}}$
(iv)Given ${a_3} = 15,\,{S_{10}} = 125,\,find\,d\,and\,{a_{10}}$
(v) Given $d = 5,\,{S_9} = 75,\,find\,a\,and\,{a_9}$
(vi) Given a = 2, d = 8 , ${S_n} = 90,\,find\,n\,and\,{a_n}$
(vii) Given a = 8, ${a_n} = 62,\,{S_n} = 210$, find n and d.
(viii) Given ${a_n} = 4,\,d = 2,\,{S_n} = - 14$, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144, and there are total 9 terms. Find a.
Sol.        (i) We have a = 5, d = 3 and ${a_n} = 50$
$\Rightarrow$ a + (n – 1)d = 50
$\Rightarrow$ 5 + (n – 1) 3 = 50
$\Rightarrow$ 3(n – 1) = 50 – 5
$\Rightarrow$ $n - 1 = {{45} \over 3} = 15$
$\Rightarrow$ n = 15 + 1 = 16
Putting n = 16, a = 5 and $\ell = {a_n} = 50\,in\,{S_n} = {n \over 2}\left( {a + \ell } \right)$
We get
${S_{16}} = {{16} \over 2}\left( {5 + 50} \right) = 8 \times 55 = 440$
Hence, n = 16 and ${S_{16}} = 440$

(ii) We have a = 7 and ${a_{13}} = 35$
Let d be the common difference of the given AP. Then,
$\Rightarrow$ ${a_{13}} = 35$
$\Rightarrow$ a + 12 d = 35
$\Rightarrow$ 7 + 12 d = 35 [Since a = 7]
$\Rightarrow$ 12d = 35 – 7 = 28
$\Rightarrow$ $d = {{28} \over {12}} = {7 \over 3}$
Putting n = 13, a = 7 and $\ell = {a_{13}} = 35\,in$
${S_n} = {n \over 2}\left( {a + \ell } \right),we\,get$
${S_{13}} = {{13} \over 2}\left( {7 + 35} \right)$
$= {{13} \over 2} \times 42$
= 13 × 21 = 273
Hence, $d = {7 \over 3}and\,{S_{13}} = 273$

(iii) We have ${a_{12}} = 37,\,d = 3$
Let a be the first term of the given AP. Then,
${a_{12}} = 37$
$\Rightarrow$ a + 11d = 37
$\Rightarrow$ a + 11(3) = 37
$\Rightarrow$ a + 11(3) = 37 [Since d = 3]
$\Rightarrow$ a = 37 – 33 = 4
Putting n = 12, a = 4 and $\ell = {a_{12}} = 37\,in\,$
${S_n} = {n \over 2}\left( {a + \ell } \right),\,we\,get$
${S_{12}} = {{12} \over 2}\left( {4 + 37} \right) = 6 \times 41 = 246$
Hence, , a = 4 and ${S_{12}} = 246$

(iv) We have, ${a_3} = 15,{S_{10}} = 125$
Let a be the first term and d the common difference of the given AP. Then,
${a_3} = 15\,\,and\,\,{S_{10}} = 125$
$\Rightarrow$ a + 2d = 15 ... (1)
and ${{10} \over 2}$ [2a + (10 – 1)d] = 125
$\Rightarrow$ 5(2a + 9d) = 125
$\Rightarrow$ 2a + 9d = 25 ... (2)
2 × (1) – (2) gives,
2(a + 2d) – (2a + 9d) = 2 × 15 – 25
$\Rightarrow$ 4d – 9d = 30 – 25
$\Rightarrow$ – 5d = 5
$\Rightarrow$ $d = - {5 \over 5} = - 1$
Now, ${a_{10}} = a + 9d = \left( {a + 2d} \right) + 7d$
= 15 + 7 (– 1) [Using (1)]
= 15 – 7 = 8
Hence, d = – 1 and ${a_{10}} = 8$

(v) We have d = 5 , ${S_9} = 75$
Let a be the first term of the given AP. Then,
${S_9} = 75$
$\Rightarrow$ ${9 \over 2}\left[ {2a + \left( {9 - 1} \right)5} \right] = 75$
$\Rightarrow$ ${9 \over 2}\left( {2a + 40} \right) = 75$
$\Rightarrow$ 9a + 180 = 75
$\Rightarrow$ 9a = 75 – 180
$\Rightarrow$ 9a = – 105
$\Rightarrow$ $a = {{ - 105} \over 9} = {{ - 35} \over 3}$
Now ${a_9} = a + 8d = {{ - 35} \over 3} + 8 \times 5$
$= {{ - 35 + 120} \over 3} = {{85} \over 3}$
Hence, $a = {{ - 35} \over 3}and\,{a_9} = {{85} \over 3}$

(vi) We have, a = 2, d = 8 , ${S_n} = 90$
${S_n} = 90$
$\Rightarrow$ ${n \over 2}\left[ {2 \times 2 + \left( {n - 1} \right)8} \right] = 90$
$\Rightarrow$ ${n \over 2}\left( {4 + 8n - 8} \right) = 90$
$\Rightarrow$ ${n \over 2}\left( {8n - 4} \right) = 90$
$\Rightarrow$ $n\left( {4n - 2} \right) = 90$
$\Rightarrow$ $4{n^2} - 2n - 90 = 0$
Therefore, $n = {{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4 \times 4 \times \left( { - 90} \right)} } \over {2 \times 4}}$
$= {{2 \pm \sqrt {4 + 1440} } \over 8}$
$= {{2 \pm \sqrt {1444} } \over 8}$
$= {{2 \pm 38} \over 8}$
$= {{40} \over 8},{{ - 36} \over 8} = 5,{{ - 9} \over 2}$
But n cannot be negative
Therefore, n = 5
Now ${a_n} = a + \left( {n - 1} \right)d$
$\Rightarrow$ ${a_5} = 2 + \left( {5 - 1} \right)8 = 2 + 32 = 34$
Hence, n = 5 and ${a_n} = 34$

(vii) We have , a = 8, ${a_n} = 62,\,{S_n} = 210$
Let d be the common difference of the given AP.
Now, ${S_n} = 210$
$\Rightarrow$ ${n \over 2}\left( {a + \ell } \right) = 210$
$\Rightarrow$ ${n \over 2}\left( {8 + 62} \right) = 210$ [Since $a = 8,\,{a_n} = 62$]
$\Rightarrow$ ${n \over 2} \times 70 = 210$
$\Rightarrow$ $n = 210 \times {2 \over {70}} = 3 \times 2 = 6$
and ${a_n} = 62$ $\Rightarrow$ ${a_6} = 62$
$\Rightarrow$ a + 5d = 62
$\Rightarrow$ 8 + 5d = 62 [since a = 8]
$\Rightarrow$ 5d = 62 – 8 = 54
$\Rightarrow$ $d = {{54} \over 5}$
Hence, $d = {{54} \over 5}$ and n = 6

(viii) We have ${a_n} = 4,\,d = 2,{S_n} = - 14$
Let a be the first term of the given AP. Then.
${a_n} = 4$
$\Rightarrow$ a + (n – 1)2 = 4 [since d = 2]
$\Rightarrow$ a = 4 – 2 (n – 1) ... (1)
and ${S_n} = - 14$
$\Rightarrow$ ${n \over 2}\left( {a + \ell } \right) = - 14$ [since $\ell = {a_n}$]
$\Rightarrow$ n (a + 4) = – 28
$\Rightarrow$ n[4 – 2 (n – 1) + 4] = – 28
$\Rightarrow$ n (4 – 2n + 2 + 4) = – 28
$\Rightarrow$ n(– 2n + 10) = – 28
$\Rightarrow$ n (– n + 5) = – 14
$\Rightarrow$ $- {n^2} + 5n = - 14$
$\Rightarrow$ ${n^2} - 5n - 14 = 0$
$\Rightarrow$ (n – 7) (n + 2) = 0
$\Rightarrow$ n = 7 or – 2
But n cannot be negative
n = 7
Putting n = 7 in (1), we get
a = 4 – 2 (7 – 1) = 4 – 2 × 6
= 4 – 12 = – 8
Hence, n = 7 and a = – 8

(ix) We have, a = 3, n = 8, S = 192
Let d be the common difference of the given AP.
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]$
$\Rightarrow$ $192 = {8 \over 2}\left[ {2 \times 3 + \left( {8 - 1} \right)d} \right]$
$\Rightarrow$ 192 = 4(6 + 7d)
$\Rightarrow$ 48 = 6 + 7d
$\Rightarrow$ 7d = 48 – 6
$\Rightarrow$ 7d = 42
$\Rightarrow$ $d = {{42} \over 7} = 6$
Hence, d = 6

(x) We have l = 28, S = 144, n = 9
Let a be the first term of the given AP.
S = 144
$\Rightarrow$ ${n \over 2}\left( {a + \ell } \right) = 144$
$\Rightarrow$ ${9 \over 2}\left( {a + 28} \right) = 144$
$\Rightarrow$ $a + 28 = 144 \times {2 \over 9}$
$\Rightarrow$ a + 28 = 32
$\Rightarrow$ a = 32 – 28 = 4
Hence, a = 4

Q.4      How many terms of the AP : 9 , 17, 25, ... must be taken to give a sum of 636 ?
Sol.        Let the first term be a = 9 and common difference d = 17 – 9 = 8. Let the sum of n terms be 636. Then,
${S_n} = 636$
$\Rightarrow$ ${n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right] = 636$
$\Rightarrow$ ${n \over 2}\left[ {2 \times 9 + \left( {n - 1} \right)8} \right] = 636$
$\Rightarrow$ ${n \over 2}\left( {18 + 8n - 8} \right) = 636$
$\Rightarrow$ ${n \over 2}\left( {8n + 10} \right) = 636$
$\Rightarrow$ n(4n + 5) = 636
$\Rightarrow$ $4{n^2} + 5n - 636 = 0$
Therefore, $n = {{ - 5 \pm \sqrt {25 - 4 \times 4 - 636} } \over {2 \times 4}}$
$= {{ - 5 \pm \sqrt {25 + 10176} } \over 8}$
$= {{ - 5 \pm \sqrt {10201} } \over 8}$
$= {{ - 5 \pm 101} \over 8} = {{96} \over 8},{{ - 106} \over 8}$
$12,{{ - 53} \over 4}$
But n cannot be negative
Therefore, n = 12
Thus, the sum of 12 terms is 636.

Q.5     The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Sol.          Let a be the first term and d the common difference of the AP such that.
a = 5, l = 45 and S = 400
Therefore, S = 400
$\Rightarrow$ ${n \over 2}\left( {a + \ell } \right) = 400$
$\Rightarrow$ $n\left( {5 + 45} \right) = 400 \times 2$
$\Rightarrow$ n(50) = 400 × 2
$\Rightarrow$ $n = {{400 \times 2} \over {50}} = 8 \times 2 = 16$
and l = 45 $\Rightarrow$ a + (n – 1) d = 45
$\Rightarrow$ 5 + (16 – 1)d = 45
$\Rightarrow$ 15d = 45 – 5 = 40
$\Rightarrow$ $a = {{40} \over {15}} = {8 \over 3}$
Hence, the number of term is 16 and the common difference is ${8 \over 3}$.

Q.6      The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Sol.         Let a be the first term and d be the common difference. Let l be its last term. Then a = 17, $\ell = {a_n} = 350$, d = 9 .
$\ell = {a_n} = 350$
$\Rightarrow$ a + (n – 1) d = 350
$\Rightarrow$ 17 + (n – 1)9 = 350
$\Rightarrow$ 9(n – 1) = 350 – 17 = 333
$\Rightarrow$ $n - 1 = {{333} \over 9} = 37$
$\Rightarrow$ n = 37 +1 = 38
Putting a = 17, l = 350, n = 38
in ${S_n} = {n \over 2}\left( {a + \ell } \right),we\,\,get$
${S_{38}} = {{38} \over 2}\left( {17 + 350} \right)$
= 19 × 367 = 6973
Hence, there are 38 terms in the AP having their sum as 6973.

Q.7      Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Sol.         Let a be the first term and d the common difference of the given AP then,
d = 7 and ${a_{22}} = 149$
$\Rightarrow$ a + (22 – 1) d = 149
$\Rightarrow$ a + 21 × 7 = 149
$\Rightarrow$ a = 149 – 147 = 2
Putting n = 22, a = 2 and d = 7 in
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]$, we get
${S_{22}} = {{22} \over 2}\left[ {2 \times 2 + \left( {22 - 1} \right)7} \right]$
= 11(4 + 21 × 7)
= 11(4 + 147)
= 11 × 151 = 1661
Hence, the sum of first 22 terms is 1661.

Q.8      Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Sol.            Let a be the first term and d the common difference of the given AP. Then,
${a_2} = 14\,\,and\,\,{a_3} = 18$
$\Rightarrow$ a + d = 14 and a + 2d = 18
Solving these equations , we get
d = 4 and a = 10
Putting a = 10, d = 4 and n = 51 in
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]$, we get
${S_{51}} = {{51} \over 2}\left[ {2 \times 10 + \left( {51 - 1} \right) \times 4} \right]$
$= {{51} \over 2}\left[ {20 + 50 \times 4} \right]$
$= {{51} \over 2}\left( {20 + 200} \right) = {{51} \over 2} \times 220$
= 51 × 110 = 5610

Q.9       If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of n terms.
Sol.         Let a be the first term and d the common difference of the given AP. Then.
${S_7} = 49\,\,and\,\,{S_{17}} = 289$
$\Rightarrow$ ${7 \over 2}\left[ {2a + \left( {7 - 1} \right)d} \right] = 49$
$\Rightarrow$ ${7 \over 2}\left( {2a + 6d} \right) = 49$
$\Rightarrow$ a + 3d = 7 ... (1)
and ${{17} \over 2}\left[ {2a + \left( {17 - 1} \right)d} \right] = 289$
$\Rightarrow$ ${{17} \over 2}\left( {2a + 16d} \right) = 289$
$\Rightarrow$ a + 8d = 17 ... (2)
Solving these two equations, we get
$\Rightarrow$ 5d = 10 , d = 2 and a = 1
Therefore, ${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]$
$= {n \over 2}\left[ {2 \times 1 + \left( {n - 1} \right)2} \right]$
$= {n \over 2}\left( {2 + 2n - 2} \right) = {n \over 2} \times 2n = {n^2}$

Q.10      Show that ${a_1},{a_2}....\,{a_n}....$ form an AP where ${a_n}$ is defined as below :
(i) ${a_n} = 3 + 4n$ (ii) ${a_n} = 9 - 5n$
Also find the sum of the first 15 term in each case.
Sol.           (i) We have, ${a_n} = 3 + 4n$
Substituting n = 1, 2, 3, 4, ... , n , we get
The sequence 7, 11, 15, 19, .... (3 + 4n) which is an AP with common difference 4.
Putting a = 7, d = 4 and n = 15 in
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right]$, we get
${S_{15}} = {{15} \over 2}\left[ {2 \times 7 + \left( {15 - 1} \right)4} \right]$
$= {{15} \over 2}\left( {14 + 14 \times 4} \right) = {{15} \over 2}\left( {14 + 56} \right)$
$= {{15} \over 2} \times 70 = 15 \times 35 = 525$

(ii) We have, ${a_n} = 9 - 5n$
Substituting n = 1, 2, 3, 4, .... n, we get
The sequence 4, – 1, – 6, – 11, .... (9 – 5n), which is an AP with common difference – 5.
Putting a = 4, d = – 5 and n = 15 in
${S_n} = {n \over 2}\left[ {2a + \left( {n - 1} \right)d} \right],$ we get
${S_{15}} = {{15} \over 2}\left[ {2 \times 4 + \left( {15 - 1} \right)\left( { - 5} \right)} \right]$
$= {{15} \over 2}\left( {8 + 14 \times - 5} \right)$
$= {{15} \over 2}\left( {8 - 70} \right) = {{15} \over 2} \times - 62$
= 15 × – 31 = – 465

Q.11     If the sum of the first n terms of an AP is 4n $- {n^2}$, what is the first term (that is ${S_1}$)? What is the sum of first two terms ? What is the second term? Similarly, find the 3rd the 10th and the nth terms.
Sol.          According to the question,
${S_n} = 4n - {n^2}$
${S_1} = 4 \times 1 - {1^2}$
= 4 – 1 = 3
$\Rightarrow$ First term = 3
Now, sum of first two terms = ${S_2} = 4 \times 2 - {2^2}$
$= 8 - 4 = 4$
Therefore Second term $= {S_2} - {S_1} = 4 - 3 = 1$
$= {S_3} = 4 \times 3 - {3^2}$
= 12 – 9 = 3
Therefore Third term = ${S_3} - {S_2}$ = 3 – 4 = – 1
${S_9} = 4 \times 9 - {9^2}$
= 36 – 81 = – 45
and, ${S_{10}} = 4 \times 10 - {10^2}$
= 40 – 100 = – 60
Therefore Tenth term = ${S_{10}} - {S_9}$
= – 60 – (– 45)
= – 60 + 45 = – 15
Also, ${S_n} = 4n - {n^2}$
and ${S_{n - 1}} = 4\left( {n - 1} \right) - {\left( {n - 1} \right)^2}$
$= 4n - 4 - {n^2} + 2n - 1$
$= - {n^2} + 6n - 5$
Therefore, nth term = ${S_n} - {S_{n - 1}}$
$= 4n - {n^2} - \left( { - {n^2} + 6n - 5} \right)$
$= 4n - {n^2} + {n^2} - 6n + 5 = 5 - 2n$

Q.12      Find the sum of the first 40 positive integers divisible by 6.
Sol.          The first positive integers divisible by 6 are 6, 12, 18, .... Clearly, it is an AP with first term a = 6 and common difference d = 6.
We want to find ${S_{10}}$
Therefore, ${S_{40}} = {{40} \over 2}\left[ {2 \times 6 + \left( {40 - 1} \right)6} \right]$
= 20 (12 + 39 × 6)
= 20(12 + 234) = 20 × 246 = 4920

Q.13      Find the sum of the first 15 multiples of 8.
Sol.           The first 15 multiples of 8 are 8 × 1, 8 × 2, 8 × 3, ... 8 × 15 i.e., 8, 16, 24 .... 120, which is an AP.
Therefore Sum of 1st 15 multiples of
$8 = {{15} \over 2}\left( {8 + 120} \right)$
$\left[ {{S_n} = {n \over 2}\left( {a + \ell } \right)} \right]$
$= {{15} \over 2} \times 128$ = 15 × 64 = 960

Q.14      Find the sum of the odd numbers between 0 and 50.
Sol.          The odd numbers between 0 and 50 are 1, 3, 5, 49. They form an AP and there are 25 terms.
Therefore, Their sum $= {{25} \over 2}\left( {1 + 49} \right)$
$= {{25} \over 2} \times 50$ = 25 × 25 = 625

Q.15     A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. how much money the contractor has to pay as penalty, if he has delayed the work by 30 days?
Sol.           Here a = 200 , d = 50 and n = 30
Therefore, $S = {{30} \over 2}\left[ {2 \times 200 + \left( {30 - 1} \right)50} \right]$
$\left[ {Since\,\,{S_n} = {n \over 2}(2a + \left( {n - 1} \right)d} \right]$
= 15(400 + 29 × 50)
= 15(400 + 1450)
= 15 × 1850
= 27750
Hence, a delay of 30 days costs the contractor Rs 27750.

Q.16      A sum of Rs 700 is to be used to give seven each prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Sol.            Let the respective prizes be a + 60, a + 40, a + 20, a, a – 20, a – 40, a – 60
Therefore, The sum of the prizes is
a + 60 + a + 40 + a + 20 + a + a – 20 + a – 40 + a – 60 = 700
$\Rightarrow$ 7a = 700
$\Rightarrow$ $a = {{700} \over 7} = 100$
Therefore, The seven prizes are 100 + 60, 100 + 40, 100 + 20, 100, 100 – 20, 100 – 40, 100 – 60
or 160, 140, 120, 100, 80, 60, 40 (in Rs)

Q.17     In a school, students thought of planting trees in an around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Sol.             Since there are three sections of each class, so the number of trees planted by class I, class II, class III,... class XII are 1 × 3, 2 × 3, 3 × 3, .... 12 × 3 respectively.
i.e., 3, 6, 9, ... 36. Clearly, it form an AP.
The sum of the number of the trees planted by these classes.
$= {{12} \over 2}\left( {3 + 36} \right) = 6 \times 39 = 234$

Q.18      A spiral is made up of successive semicircles , with centres alternately at A and B, starting with cenre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in fig. What is the total length of such a spiral made up of thirteen consecutive semicircles ?
$\left( {Take\,\pi = {{22} \over 7}} \right)$
Sol.       Length of a semi-circum ference = $\pi r$ where r is the radius of the circle.
Therefore, Length of spiral made up of thirteen consecutive semicircles.
$= \left( {\pi \times 0.5 + \pi \times 1.0 + \pi + 1.5 + \pi \times 2.0 + .... + \pi \times 6.5} \right)cm$
$= \pi \times 0.5\left( {1 + 2 + 3 + .... + 13} \right)cm$
$= \pi \times 0.5 \times {{13} \over 2}\left( {2 \times 1 + [13 - 1} \right) \times 1]\,cm$
$= {{22} \over 7} \times {5 \over {10}} \times {{13} \over 2} \times 14\,cm = 143\,cm$

Q.19      200 logs are stacked in the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Sol.       Clearly logs stacked in each row form a sequence 20 + 19 + 18 + 17 + .... It is an AP with a = 20, d = 19 – 20 = – 1.
Let ${S_n} = 200$. Then
${n \over 2}\left[ {2 \times 20 + \left( {n - 1} \right)\left( { - 1} \right)} \right] = 200$
$\Rightarrow$ n(40 – n + 1) = 400
$\Rightarrow$ ${n^2} - 41n + 400 = 0$
$\Rightarrow$ $\left( {n - 16} \right)\left( {n - 25} \right) = 0$
$\Rightarrow$ n = 16 or 25
Here the common difference is negative.
The terms go on diminishing and 21st term becomes zero. All terms after 21st term are negative. These
negative terms when added to positive terms from 17th term to 20th term, cancel out each other and the sum remains the same.

Thus n = 25 is not valid for this problem. So we take n = 16.
Thus, 200 logs are placed in 16 rows.
Number of logs in the 16th row
$= {a_{16}}$
= a + 15d
= 20 + 15(–1)
= 20 – 15 = 5

Q.20      In a potato race, a bucket is placed at the starting point, which is 5 cm from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure).

A competitor starts from the bucket, picks up the earest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
Sol.      To pick up the first potato second potato, third potato, fourth potato, ....
The distance (in metres) run by the competitor are
2 × 5 ; 2 × (5 + 3), 2 × (5 + 3 + 3), 2 × (5 + 3 + 3 + 3), ....
i.e., 10, 16, 22, 28, ....
which is in AP with a = 10, d = 16 – 10 = 6
Therefore, The sum of first ten terms,
${S_{10}} = {{10} \over 2}\left[ {(2 \times 10 + \left( {10 - 1} \right) \times 6)} \right]$
= 5(20 + 54) = 5 × 74 = 370
Therefore, The total distance the competitor has to run is 370 m.

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