The fifth term of an A.P. is 1 whereas its $31^{s t}$ term is -77. Find sum of how many terms will be 20.

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Solution

$T_{5} = a + 4 d = 1,$ $T_{31} = a + 30 d = - 77$
Solving the above two, we get
$a = 13$ and $d = - 3$
$S_{15} = [n / 2] [2 a + [n - 1] d] = [15 / 2] [26 + 14 [- 3]] = - 120$
Let $T_{n} = - 17$ Then $a + [n - 1] d = - 17$
$13 + [n - 1] [- 3] = 17$
$3 n = 33$ or $n = 11$
Let $S_{n} = 20$ Then $(n / 2) [2 a + (n - 1) d] = 20$
$3 n^{2} - 29 n + 40 = 0$
$[n - 8] [3 n - 5] = 0$
$n = 8$ . The $d$ value of $n$ can not be fractional.

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31

st term is

- 77

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(i) How many terms of the sequence $18, 16, 14, . . .$ should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are $- 14$ and $2$ respectively and the sum of the terms is $40$ ?
(iii) How many terms of the A.P. $9, 17, 25, . . .$ must be taken so that their sum is $636$ ?
(iv) How many terms of the A.P. $63, 60, 57, . . .$ must be taken so that their sum is $693$ ?
(v) How many terms of the A.P. $27, 24, 21...$ should be taken so that their sum is zero?

(i) How many terms of the sequence $18, 16, 14, . . .$ should be taken so that their sum is zero?
(ii) How many terms are there in the A.P. whose first and fifth terms are $- 14$ and $2$ respectively and the sum of the terms is 40?
(iii) How many terms of the A.P., $9, 17, 25, . . .$ must be taken so that their sum is $636$ ?
(iv) How many terms of the A.P., $63, 60, 57, . . .$ must be taken so that their sum is $693$ ?
(v) How many terms of the A.P., $27, 24, 21...$ should be taken so that their sum is zero?
(vi) How many terms of the A.P., $45, 39, 33...$ must be taken so that their sum is $180$ ?

Q. (i) How many terms of the sequence 18, 16, 14, ... should be taken so that their sum is zero?

(ii) How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

(iii) How many terms of the A.P. 9, 17, 25,... must be taken so that their sum is 636?

(iv) How many terms of the A.P. 63, 60, 57, ... must be taken so that their sum is 693?

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The fifth term of an A.P. is 1 whereas its 31st term is -77. Find sum of how many terms will be 20.

T5=a+4d=1, T31=a+30d=−77Solving the above two, we geta=13 and d=−3S15=[n/2][2a+[n−1]d]=[15/2][26+14[−3]]=−120Let Tn=−17 Then a+[n−1]d=−1713+[n−1][−3]=173n=33 or n=11Let Sn=20 Then (n/2)[2a+(n−1)d]=20 3n2−29n+40=0[n−8][3n−5]=0n=8 . The d value of n can not be fractional.

The fifth term of an A.P. is 1 whereas its $31^{s t}$ term is -77. Find sum of how many terms will be 20.