The 26^{th}, 11^{th} and the last term of an AP are 0, 3 and `- 1/5`, respectively. Find the common difference and the number of terms.

#### Solution

Let the first term, common difference and number of terms of an AP are a, d and n, respectively.

We know that, if last term of an AP is known, then

l = 8 + (11 – 1)d ........(i)

And n^{th} term of an AP is

T_{n} = a + (n – 1)d .........(ii)

Given that, 26^{th} term of an AP = 0

⇒ T_{26} = a + (26 – 1 )d = 0 ......[From equation (i)]

⇒ 8 + 25d = 0 ........(iii)

11^{th} term of an AP = 3

⇒ T_{11} = s + (11 – 1)d = 3 .......[From equation (ii)]

⇒ 8 + 10d = 3 ........(iv)

And last term of an AP = `-1/5`

⇒ l = a + (n – 1 )d .......[From equation (i)]

⇒ `-1/5` = a + (n – 1 )d ........(v)

Now, subtracting equation (iv) from equation (iii),

15d = – 3

⇒ d = `- 1/5`

Put the value of d in equation (iii), we get

`a + 25(-1/5)` = 0

⇒ a – 5 = 0

⇒ 8 = 5

Now put the value of a, d in equation (v), we get

`-1/5 = 5 + (n - 1)(-1/5)`

⇒ – 1 = 25 – (n – 1)

⇒ – 1 = 25 – n + 1

⇒ n = 25 + 2 = 27

Hence, the common difference and number of terms are `-1/5` and 27, respectively.