Question

3) In A.P. 19" term is 52 and 38 term is 128, find sum of first 56 terms.​

Answer 1

Answer:

5040 is the sum of 56 terms of following AP.

Step-by-step explanation:

According to first statement

19th term of an AP is 52 .

tn = a + (n - 1 ) d .

t19 = a + (19 - 1) d.

52 = a + 18d .

52 - 18d = a. --------- [EQ 1]

From second statement

38th term of an AP is 128

t38 = a + (38 - 1 ) d

128 = a + 37d. ---------- [EQ 2]

substitute value of a from equation 1 in equation 2.

128 = a + 37d

128 = 52 - 18d + 37d. [ substitute a from eq 1]

128 - 52 = 37d - 18d

76 = 19d

76 ÷ 19 = d

4 = d

So we have got value of "d" that is 4.

Put value of d in equation 1

52 - 18d = a

52 - 18(4) = a

52 - 72 = a

-20 = a

Value of a is -20.

formula to find sum of n terms is

Sn = n ÷ 2 × [ 2a - (n-1) d ]

S56 = 56 ÷ 2 × [ 2 × -20 +( 56 - 1 ) × 4 ]

S56 = 28 × [ -40 + (55) × 4 ]

S56 = 28 × [ -40 + 220 ]

S56 = 28 × [ 180]

S56 = 5040.

Sum of 56 terms of AP is 5040.