Question

In an A.P the first term is 2, the last term is 29, the sum is 155. Find the common difference of this A.P.

Answer 1

Hint: In this question directly use the formula for sum of n terms of an A.P but terms of first and last term that is ${S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$ and the formula for nth term of an A.P which is ${a_n} = {a_1} + \left( {n - 1} \right)d$, to get the answer.

Complete step-by-step answer:
Given data
In an A.P first term (a₁ = 2), last term (a_n = 29), sum (S_n = 155).
We have to find out the value of common difference (d).
Now as we know that the formula of sum of n term of an A.P is
$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {a + {a_n}} \right)$ Where symbols have their usual meaning and (n) is the number of terms.
So substitute the values in this equation we have,
$ \Rightarrow 155 = \dfrac{n}{2}\left( {2 + 29} \right)$
$ \Rightarrow n = \dfrac{{155 \times 2}}{{31}} = 10$
So the number of terms in an A.P is 10.
Now we all know the formula of n^th term or last term of an A.P is
$ \Rightarrow {a_n} = {a_1} + \left( {n - 1} \right)d$ Where symbols have their usual meanings.

Now substitute the values in this equation we have,
$ \Rightarrow 29 = 2 + \left( {10 - 1} \right)d$
Now simplify the above equation we have,
$ \Rightarrow 9d = 29 - 2 = 27$
Now divide by 9 we have,
$ \Rightarrow d = \dfrac{{27}}{9} = 3$
So the value of common difference of an A.P is 3.
So this is the required answer.

Note: Numbers in a series are said to be in A.P if and only if the common difference that is the difference between two consecutive terms is constant. The sum of n terms of an A.P can also be written as ${S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)$. While finding n that is the number of terms in an A.P we need to take care of one thing that n can’t be negative.