Question

nes
3 a) Find the sum of “n” terms of the following progression
7, 14, 28, 56, .... (where n = 10)
OR
b) Find the sum of “n" terms of the following series
4, 7, 11, 14,..... (where n = 10).
- prob​

Answer 1

Answer:

a) n/2 ( 2a+(n-1)d)

10/2(2(7)+(10-1)7)

5(14+63)

5(77)

385

b) this is not AP because D is not same

4-7=-3

7-11=-4

-3 is not equal to -4

so if it is not an AP then we can't find its sum.

thanks.................

Answer 2

Answer:

given ,

• a = 7
• n = 10
• d = 7

we know that,

• Sum = n/2 [ 2a + (n-1)d ]

= 10/2 [2*7 + (10-1)7 ]

= 5 (14 + 9*7)

= 5 (14 + 63)

= 5 * 77

= 385.

therefore the sum of 10th term of ap is 385