In an A.p., the sum of first n terms is P , the sum of the next n terms is Q and the sum of further next n terms is R. Show that P, Q, R is an A.P.


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Question Number 46805 by 786786AM last updated on 31/Oct/18

$In an A . p ., the sum of first n terms is P, the sum of the next n terms is Q and$ $the sum of further next n terms is R . Show that P, Q, R is an A . P .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18
	$first term is a and common difference is d P=(n/2)[2a+(n−1)d].....(1) P+Q=((2n)/2)[2a+(2n−1)d] P+Q+R=((3n)/2)[2a+(3n−1)d] Q=((2n)/2)[2a+(2n−1)d]−(n/2)[2a+(n−1)d] =(n/2)[4a+4nd−2d−2a−nd+d] =(n/2)[2a+(3n−1)d] Q−P =(n/2)[2a+(3n−1)d−2a−(n−1)d] =(n/2)[(3n−1−n+1)d =n^2 d R=P+Q+R−(P+Q) R=((3n)/2)[2a+(3n−1)d]−((2n)/2)[{2a+(2n−1d)] R=(n/2)[6a+9nd−3d−4a−4nd+2d] R=(n/2)[2a+(5n−1)d] R−Q =(n/2)[2a+(5n−1)d]−(n/2)[2a+(3n−1)d] =(n/2)[(5n−1−3n+1)d] =(n/2)×2nd =n^2 d Q−P=R−Q so P,Q,R in A.P$
	$f i r s t t e r m i s a a n d c o m m o n d i f f e r e n c e i s d$ $P = \frac{n}{2} [2 a + (n - 1) d] . . . . . (1)$ $P + Q = \frac{2 n}{2} [2 a + (2 n - 1) d]$ $P + Q + R = \frac{3 n}{2} [2 a + (3 n - 1) d]$ $Q = \frac{2 n}{2} [2 a + (2 n - 1) d] - \frac{n}{2} [2 a + (n - 1) d]$ $= \frac{n}{2} [4 a + 4 n d - 2 d - 2 a - n d + d]$ $= \frac{n}{2} [2 a + (3 n - 1) d]$ $Q - P$ $= \frac{n}{2} [2 a + (3 n - 1) d - 2 a - (n - 1) d]$ $= \frac{n}{2} [(3 n - 1 - n + 1) d$ $= n^{2} d$ $R = P + Q + R - (P + Q)$ $R = \frac{3 n}{2} [2 a + (3 n - 1) d] - \frac{2 n}{2} [{2 a + (2 n - 1 d)]$ $R = \frac{n}{2} [6 a + 9 n d - 3 d - 4 a - 4 n d + 2 d]$ $R = \frac{n}{2} [2 a + (5 n - 1) d]$ $R - Q$ $= \frac{n}{2} [2 a + (5 n - 1) d] - \frac{n}{2} [2 a + (3 n - 1) d]$ $= \frac{n}{2} [(5 n - 1 - 3 n + 1) d]$ $= \frac{n}{2} \times 2 n d$ $= n^{2} d$ $Q - P = R - Q$ $s o P, Q, R i n A . P$
	Commented by 786786AM last updated on 01/Nov/18

	$Tk yu sir .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18

	$m o s t w e l c o m e . . .$
	Answered by MrW3 last updated on 01/Nov/18

	$l e t d = c o m m o n d i f f e r e n c e$ $P = a_{1} + a_{2} + . . . + a_{n}$ $Q = a_{n + 1} + a_{n + 2} + . . . + a_{2 n} = (a_{1} + n d) + (a_{2} + n d) + . . . + (a_{n} + n d)$ $= (a_{1} + a_{2} + . . . + a_{n}) + n \times n d$ $= P + n^{2} d$ $\Rightarrow P - Q = n^{2} d$ $s i m i l a r l y$ $R = a_{2 n + 1} + a_{2 n + 2} + . . . + a_{3 n} = Q + n^{2} d$ $\Rightarrow R - Q = n^{2} d$ $s i n c e R - Q = Q - P$ $\Rightarrow P, Q, R a r e i n A . P .$
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