Question

the sum of first 13 terms of a.p. is 21 and the sum of first 21 terms is 13 . find the sum of the first 34 terms.

Answer 1

S13=21
s21=13
sm=sn....so sm+n=0
s34=0

Answer 2

Answer

Given :- Sum of 13 terms = 21

Sum of 21 terms = 13

Sum of terms = n/2 [2a + (n-1) d]

S 13 = 13/2 [ 2a + (13-1) d] = 21

13/2 [2a + 12d] = 21

26a + 156d = 42

13a + 78d = 21. --------(1)

S 21 = 21/2 [ 2a + (21-1) d ] = 13

21/2 [ 2a + 20d ] = 13

42a + 420d = 26

21a + 210d = 13. ---------(2)

Multipling (1) by 21 and (2) by 13, we get

273a + 1638d = 441. ------(3)

273a + 2730d = 169. ------(4)

Subtracting (3) and (4) ,we get

d = -68/273. -------(5)

Put the value of d in (2)

21a + 210(-68/273) = 13

21a - 680/13 = 13

21a = 13 + 680/13

21a = 849/13

a = 849/13*21

a = 849/273

a = 283/91. ------(6)

Now sum of first 34 terms

S 34 = 34/2 [2 (283/91) + (34-1)( -68/273]

= 17 [ 566/91 - 33(-68/273]

= 17 [ 566/91 - 2244/273 ]

= 17 [ 566/91 - 748/91 ]

= 17 [ 566-748/91 ]

= 17 [ -182/91 ]

= -3094/91

= -34

So the sum of first 34 terms of the AP = -34.