Question

The 26th 11th and last term of an ap are 0,3 & -15. Find common difference and number of terms.

Answer 1

We will use the formula
an = a + (n - 1)d
a26 = 0
a26 = a + (n - 1)d = 0
a26 = a + (26 - 1)d = 0
a + 25d = 0 ............. (1)

a11 = a + (n - 1)d = 3
a11 = a + (11 - 1)d = 3
a + 10d = 3 ................. (2)

Subtract (1) from (2) we get,
(a - a) + d(10 - 25) = 3 - 0
- 15d = 3
d = - 3 / 15
d = - 1 / 5

Subsitute d value in (1) a + 25d = 0 we get,
a + [(25 x ( - 1/5)] = 0
a - 5 = 0
a = 5

Subsitute both a & d value in
an = a + (n - 1)d = - 15
(5) + (n - 1) (- 1/5) = - 15
- (n - 1) / 5 = - 15 - 5
- (n - 1) / 5 = - 20
- (n - 1) = - 100
(n - 1) = 100
n - 1 = 100
n = 100 + 1
n = 101

Therefore, the difference (d) = - 1/5 & number of terms (n) = 101