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Standard XII

Mathematics

nth Term of A.P

Let there be ...

Question

Let there be an A.P with first term $a$ , common difference $d$ . If $a_{n}$ denotes its $n$ th term and $S_{n}$ the sum of first $n$ terms, find
$a$ , if $a_{n} = 28$ , $S_{n} = 144$ and $n = 9$

a = 4

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a = - 4

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a = 6

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a = - 6

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Solution

The correct option is A $a = 4$
$n = 9$

$a_{n} = a_{9} = 28$

$a + 8 d = 28$ (1)

$S_{9} = 144$

$\frac{9}{2} [2 a + 8 d] = 144$

$2 a + 8 d = 32$ (2)
Solving equation 1 and 2, we get, $a = 4$ .

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Q. Let there be an A.P. with first term 'a', common difference 'd'. If a_n denotes in n^th term and S_n the sum of first n terms, find.

(i) n and S_n, if a = 5, d = 3 and a_n = 50.

(ii) n and a, if a_n = 4, d = 2 and S_n= −14.

(iii) d, if a = 3, n = 8 and S_n = 192.

(iv) a, if a_n = 28, S_n = 144 and n= 9.

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(vi) n and a_n, if a= 2, d = 8 and S_n = 90.

Let there be an A.P. with first term 'a', common difference 'd'. If $a_{n}$ denotes its nth term and $S_{n}$ the sum of first n terms, find.
(i) n and $S_{a}$ , if a=5, d=3 and $a_{n}$ =50.
(ii) n and a, if $a_{n}$ =4, d=2 and $S_{n}$ =-14.
(iii) d, if a=3, n=8 and $S_{n}$ =192.
(iv)a, if $a_{n}$ =28, $S_{n}$ =144 and n=9.
(v) n and d, if a=8, $a_{n}$ =62 and $S_{n}$ =210.
(vi) n and $a_{n}$ , if a = 2, d=8 and $S_{n}$ =90.
(vii) k, if $S_{n} = 3 n^{2}$ +5n and $a_{k}$ =164.
(viii) $S_{22}$ , if d =22 and $a_{22}$ =149

Q. If

S_{n}

denotes the sum of

n

terms of an

A P

whose common difference is

d

and first term is

a

.
Find

S_{n} - 2 S_{n} + S_{n + 2}

Let $S_{n}$ denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by $d = S_{n} - k S_{n - 1} + S_{n - 2}$ , then k =

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Let there be an A.P with first term a, common difference d. If an denotes its nth term and Sn the sum of first n terms, finda, if an=28, Sn=144 and n=9

The correct option is A a=4n=9an=a9=28a+8d=28 (1)S9=14492[2a+8d]=1442a+8d=32 (2)Solving equation 1 and 2, we get, a=4.

Let there be an A.P with first term $a$ , common difference $d$ . If $a_{n}$ denotes its $n$ th term and $S_{n}$ the sum of first $n$ terms, find
$a$ , if $a_{n} = 28$ , $S_{n} = 144$ and $n = 9$

The correct option is A $a = 4$
$n = 9$

$a_{n} = a_{9} = 28$

$a + 8 d = 28$ (1)

$S_{9} = 144$

$\frac{9}{2} [2 a + 8 d] = 144$

$2 a + 8 d = 32$ (2)
Solving equation 1 and 2, we get, $a = 4$ .