If S n - n 2 p and S m 3- m 2 p - m - n- in an A-P- prove that S p - p 3-

Let a be the first term of the A.P. and d is the common difference. Then S_n = n^2p S_n = n/2(2a+(n 1)d) n^2p=n/2(2a+(n 1)d) np = 1/2[2a+(n 1)d] 2np =2a ...

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Byju's Answer

Standard X

Mathematics

Formula for Sum of n Terms of an AP

If S n = n 2 ...

Question

If $S_{n} = n^{2} p$ and $S_{m_{3}} = m^{2} p, m \neq n$ , in an A.P., prove that $S_{p} = p^{3}$ .

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Solution

Let a be the first term of the A.P. and d is the common difference. Then
$S_{n} = n^{2} p S_{n} = \frac{n}{2} (2 a + (n - 1) d) n^{2} p = \frac{n}{2} (2 a + (n - 1) d) n p = \frac{1}{2} [2 a + (n - 1) d] 2 n p = 2 a + (n - 1) d \dots \dots (i)$
Again,
$S_{m} = m^{2} p S_{m} = \frac{m}{2} (2 a + (m - 1) d) m p = \frac{1}{2} [2 a + (m - 1) d] 2 m p = 2 a + (m - 1) d \dots \dots (i i)$
Now subtract (1) from (2)
$2 p (m - n) = (m - n) d d = 2 p \dots \dots (i i i)$
Substitute the value of (iii) in (ii)
$\Rightarrow 2 m p = 2 a + (m - 1) 2 p \Rightarrow 2 a = 2 p \Rightarrow a = p$
The sum up to p terms will be
$S_{p} = \frac{p}{2} (2 a + (p - 1) d) \Rightarrow S_{p} = \frac{p}{2} (2 a + (p - 1) 2 p) \Rightarrow S_{p} = \frac{p}{2} (2 a + 2 p^{2} - 2 p) \Rightarrow S_{p} = p^{3}$

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