Question

The sum of the first four terms of an A.P. is 66. The sum of the last four terms is 112. If its first term is 11, then find the number of terms

Answer 1

Answer:

11

Step-by-step explanation:

Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + (n – 1) d] = 4a + (4n – 10) d

According to question -

4a + 6d = 66

⇒ 4(11) + 6d = 66 [∵ a = 11 (given)]

⇒ 6d = 12

⇒ d = 2

and,

4a + (4n –10) d = 112

⇒ 4(11) + (4n – 10)2 = 112

⇒ (4n – 10)2 = 68

⇒ 4n – 10 = 34

⇒ 4n = 44

⇒ n = 11

Thus, the number of terms of the A.P. is 11.

JAI SHREE KRISHNA

Answer 2

Let the AP is a , a + d , a + 2d , a + 3d ........

Here,

sum of first four terms = 56

a + (a + d) + (a + 2d) + (a + 3d) = 56

4a + 6d = 56

2a + 3d = 28

Given, first term ( a ) = 11

so, 2 × 11 + 3d = 28

3d = 28 - 22 = 6

d = 2

Let last term is Tₙ.

then sum of last four terms = 112

e.g. Tₙ + Tₙ_₁ + Tₙ_₂ + Tₙ_₃ = 112

use formula ,

Tₙ = a + (n - 1)d

a + (n -1)d + a + (n - 1 - 1)d + a + (n -2 - 1)d + a + (n - 3 - 1)d = 112

4a + (n - 1 + n - 2 + n - 3 + n - 4)d = 112

4a + (4n - 10)d = 112

put a = 11 and d = 2

4 × 11 + (4n - 10) × 2 = 112

44 + (4n - 10) × 2 = 112

2 × (4n - 10) = 68

4n = 44

n = 11