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-

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

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] \dots . . (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 \dots

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} + \dots + a_{n}

Q = a_{n + 1} + a_{n + 2} + \dots + a_{2 n} = (a_{1} + n d) + (a_{2} + n d) + \dots + (a_{n} + n d)

= (a_{1} + a_{2} + \dots + 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} + \dots + 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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