Question

Find the sum of 32 terms of an A.P. whose third term is 1 and the 6th term is -11.​

Answer 1

Answer:

n = 32

3rd term = a+2d=1 ------1.

6th term = a+5d= -11 ------2.

Subtracting 1 from 2

 a + 5d = -11

 a + 2d =   1

-    -          -    

       3d = -12

       d = -12/3

         d = -4

Put value of d in 1.

a+2X-4 = 1

a - 8 = 1

a = 1+8

a = 9

Now: Sum of 32 terms = 32/2[9 + (32-1)-4]

                                      = 16 [9 + 31 X -4]

                                      = 16 (9 - 124)

                                      = 16 X -115 =   - 1840

Step-by-step explanation:

Hope it helps :)

Answer 2

Step-by-step explanation:

Answer will be 1840 .

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