Question

In an A.P, the first term is 2 and the sum of first five terms is 5, then the ${\text{3}}{{\text{1}}^{st}}$term is:
${\text{a}}{\text{. 13}} \\\ {\text{b}}{\text{. 17}} \\\ {\text{c}}{\text{. - 13}} \\\ {\text{d}}{\text{. }}\dfrac{{27}}{2} \\\ {\text{e}}{\text{. - }}\dfrac{{27}}{2} \\\$

Answer 1

Hint: - ${n^{th}}$term of an A.P is given as$\left( {{a_n} = {a_1} + \left( {n - 1} \right)d} \right)$, (where d is the common difference)

Given data:
First term of an A.P$\left( {{a_1}} \right) = 2$……………… (1)
Sum of first five terms $\left( {{S_5}} \right) = 5$……………… (2)
Then we have to find out the value of ${\text{3}}{{\text{1}}^{st}}$term.
Now, we know that the sum of an A.P is
${{\text{S}}_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, (where d is the common difference)
So, ${{\text{S}}_5} = \dfrac{5}{2}\left( {2{a_1} + \left( {5 - 1} \right)d} \right)$
Now from equation (1) and (2) we have
${{\text{S}}_5} = \dfrac{5}{2}\left( {2{a_1} + \left( {5 - 1} \right)d} \right) \\\ \Rightarrow 5 = \dfrac{5}{2}\left( {2 \times 2 + 4d} \right) \\\ \Rightarrow 2 = 4 + 4d \\\ \Rightarrow d = \dfrac{{2 - 4}}{4} = \dfrac{{ - 2}}{4} = \dfrac{{ - 1}}{2} \\\$
Now, we have to find out the value of ${\text{3}}{{\text{1}}^{st}}$term.
As we know that the ${n^{th}}$term of an A.P is given as$\left( {{a_n} = {a_1} + \left( {n - 1} \right)d} \right)$
$\Rightarrow {31^{th}}$Term of the A.P is
$\Rightarrow {a_{31}} = 2 + \left( {31 - 1} \right)\left( {\dfrac{{ - 1}}{2}} \right) = \left( {2 - 15} \right) = - 13$
So, option (c) is correct.

Note: - In such types of questions the key concept we have to remember is that always remember all the general formulas of A.P which is stated above, then first find out the value of common difference using the formula of sum of an A.P then using the formula of ${n^{th}}$term of an A.P calculate the value of ${31^{th}}$term which is the required answer.