If a and d are the first term and the common difference of an AP, $S_{n}$ denotes the sum to n terms of an AP. If $S_{n} = P n + Q n^{2}$ , what are P and Q?

a = P + Q ; d = 2Q

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a = P - Q ; d = 2Q

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a = P + Q ; d = Q/2

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a = P + Q ; d = Q

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Solution

The correct option is A
a = P + Q ; d = 2Q

$S_{n}$ = ( $\frac{n}{2}$ )(2a + (n - 1)d) = na + $\frac{n (n - 1) d}{2}$ = na + $\frac{n^{2} d - n d}{2}$ = n(a - $\frac{d}{2}$ ) + ( $\frac{d}{2}$ ) $n^{2}$

Therefore,

P = $a - \frac{d}{2}$ and Q = $\frac{d}{2}$

=> a = P + Q and d = 2Q

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Similar questions

If $a$ and $d$ are the first term and the common difference of an AP, $S_{n}$ denotes the sum of $n$ terms of an AP. If $S_{n} = P n + Q n^{2}$ , what are $a$ and $d$ in terms of $P$ and $Q$ ?

Q. If

S_{n} = n P + \frac{n}{2} (n - 1)

Q, where

S_{n}

denotes the sum of first n terms of an AP, the common difference is

Q. In an AP. S_p = q, S_q = p and S_r denotes the sum of first r terms. Then, S_p_+q is equal to

(a) 0

(b) −(p + q)

(c) p + q

(d) pq

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If a and d are the first term and the common difference of an AP, Sn denotes the sum to n terms of an AP. If Sn=Pn+Qn2, what are P and Q?

The correct option is A a = P + Q ; d = 2Q Sn = ( n2)(2a + (n - 1)d) = na + n(n−1)d2 = na + n2d−nd2 = n(a - d2) + ( d2) n2 Therefore, P = a−d2 and Q = d2 => a = P + Q and d = 2Q

If a and d are the first term and the common difference of an AP, $S_{n}$ denotes the sum to n terms of an AP. If $S_{n} = P n + Q n^{2}$ , what are P and Q?

The correct option is A
a = P + Q ; d = 2Q

$S_{n}$ = ( $\frac{n}{2}$ )(2a + (n - 1)d) = na + $\frac{n (n - 1) d}{2}$ = na + $\frac{n^{2} d - n d}{2}$ = n(a - $\frac{d}{2}$ ) + ( $\frac{d}{2}$ ) $n^{2}$

Therefore,

P = $a - \frac{d}{2}$ and Q = $\frac{d}{2}$

=> a = P + Q and d = 2Q