Question

ONLY FOR CLASS 10th.....................

Answer 1

I will only tell u steps

First find sum of first 10 terms. Now n = 50

To find last 15 terms

50 - 15= 45

Find sum of first 45 terms.

Then u get 2 eqns

Solve them and u will get ur answer

Answer 2

Consider a and d as the first term and the common difference of an A.P. respectively.

n th term of an A.P:

an = a + ( n – 1)d

Sum of n terms of an A.P:

S n = n/ 2 [2a + (n – 1)d]

Given that the sum of the first 10 terms is 210.

⇒ 10 / 2 [2a + 9d ] = 210

⇒ 5[ 2a + 9 d ] = 210

⇒2a + 9d = 42 ----------- (1)

15 th term from the last = ( 50 – 15 + 1 )

th = 36 th term from the beginning

⇒ a36 = a + 35d

Sum of the last 15 terms:

15/2 [2a36 + ( 15 – 1)d ] = 2565

⇒ 15 / 2 [ 2(a + 35d) + 14d ] = 2565

⇒ 15 [ a + 35d + 7d ] = 2565

⇒a + 42d = 171 ----------(2)

From (1) and (2), we have d = 4 and a = 3.

Therefore, the terms in A.P. are 3, 7, 11, 15 . . . and 199.

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