Given:
19th term of an AP = 52
38th term = 128
Find:
the sum of the first 56 terms
Solution:
We know that,
nth term of an AP – a(n) = a + (n - 1)d
Hence,
→ a + (19 - 1)d = 52
→ a + 18d = 52 -- equation (1)
Similarly,
→ a + (38 - 1)d = 128
→ a + 37d = 128 -- equation (2)
Subtract equation (1) from (2).
→ a + 37d - (a + 18d) = 128 - 52
→ a + 37d - a - 18d = 76
→ 19d = 76
→ d = 76/19
→ d = 4
Substitute the value of d in equation (1).
→ a + 18 * 4 = 52
→ a = 52 - 72
→ a = - 20
We know,
Sum of first n terms of an AP – S(n) = n/2 * [ 2a + (n - 1)d ]
Hence,
→ S(56) = 56/2 * [ 2 (- 20) + (56 - 1) * (4) ]
→ S(56) = 28 * [ - 40 + 55 * 4 ]
→ S(56) = 28 * [ - 40 + 220 ]
→ S(56) = 28 * (180)
→ S(56) = 5040
Hence, the sum of first 56 terms of the given AP is 5040.
I hope it will help you.
Regards.
