Question

In an A.P., the first term is $2$ and the sum of the first five terms is one-fourth of the next five terms. Show that ${20^{th}}$ term is $ - 112$.

Answer 1

Hint: Assume common difference to be d and use the formula for ${{\text{n}}^{{\text{th}}}}$ term
$ \Rightarrow {{\text{T}}_n} = \left( {{\text{a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right)$ where a is the first term, d is the common difference to find the first ten terms and solve the obtained equation for d. Then use the same formula${{\text{T}}_n} = \left( {{\text{a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right)$ to find ${20^{th}}$ term.

Complete step-by-step answer:
Given, in an A.P. series the first term a=$2$
Also the sum of first five terms is one-fourth of the next five terms.
We have to show that ${20^{th}}$ term is$ - 112$.
Now we have to find the first ten terms.
We know that in A.P. the ${{\text{n}}^{{\text{th}}}}$ term is written as,
$ \Rightarrow {{\text{T}}_n} = \left( {{\text{a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right)$ where a is the first term, d is the common difference.
So we can write the first term as-
$ \Rightarrow {{\text{T}}_1} = \left( {{\text{a + }}\left( {{\text{1 - 1}}} \right){\text{d}}} \right)$
On solving we get,
$ \Rightarrow {{\text{T}}_1} = {\text{a}}$
Then, second term can be written as-
$ \Rightarrow {{\text{T}}_2} = \left( {{\text{a + }}\left( {{\text{2 - 1}}} \right){\text{d}}} \right)$
On solving we get,
$ \Rightarrow {{\text{T}}_2} = \left( {{\text{a + d}}} \right)$
Now we can write third term as-
$ \Rightarrow {{\text{T}}_3} = \left( {{\text{a + }}\left( {{\text{3 - 1}}} \right){\text{d}}} \right)$
On solving we get,
$ \Rightarrow {{\text{T}}_3} = \left( {{\text{a + 2d}}} \right)$
So following this pattern we can write the other terms as-
$ \Rightarrow {{\text{T}}_4} = \left( {{\text{a + 3d}}} \right)$
$ \Rightarrow {{\text{T}}_5} = \left( {{\text{a + 4d}}} \right)$
$ \Rightarrow {{\text{T}}_6} = \left( {{\text{a + 5d}}} \right)$
$ \Rightarrow {{\text{T}}_7} = \left( {{\text{a + 6d}}} \right)$
$ \Rightarrow {{\text{T}}_8} = \left( {{\text{a + 7d}}} \right)$
$ \Rightarrow {{\text{T}}_9} = \left( {{\text{a + 8d}}} \right)$
And $ \Rightarrow {{\text{T}}_{10}} = \left( {{\text{a + 9d}}} \right)$
On adding the first five terms we get,
$ \Rightarrow {{\text{T}}_1}{\text{ + }}{{\text{T}}_2}{\text{ + }}{{\text{T}}_3} + {{\text{T}}_4} + {{\text{T}}_5} = {\text{a + }}\left( {{\text{a + d}}} \right) + \left( {{\text{a + 2d}}} \right) + \left( {{\text{a + 3d}}} \right) + \left( {{\text{a + 4d}}} \right)$
On taking same terms together we get,
$ \Rightarrow {{\text{T}}_1}{\text{ + }}{{\text{T}}_2}{\text{ + }}{{\text{T}}_3} + {{\text{T}}_4} + {{\text{T}}_5} = {\text{a + a + a + a + a + d + 2d + 3d + 4d}}$
On solving we get,
$ \Rightarrow {{\text{T}}_1}{\text{ + }}{{\text{T}}_2}{\text{ + }}{{\text{T}}_3} + {{\text{T}}_4} + {{\text{T}}_5} = 5{\text{a + 10d}}$ --- (i)
Now on adding the next five terms, we get-
$ \Rightarrow {{\text{T}}_6}{\text{ + }}{{\text{T}}_7}{\text{ + }}{{\text{T}}_8} + {{\text{T}}_9} + {{\text{T}}_{10}} = \left( {{\text{a + 5d}}} \right) + \left( {{\text{a + 6d}}} \right) + \left( {{\text{a + 7d}}} \right) + \left( {{\text{a + 8d}}} \right) + \left( {{\text{a + 9d}}} \right)$
On solving we get,
$ \Rightarrow {{\text{T}}_6}{\text{ + }}{{\text{T}}_7}{\text{ + }}{{\text{T}}_8} + {{\text{T}}_9} + {{\text{T}}_{10}} = {\text{5a + 35d}}$ --- (ii)
Now according to question-
$ \Rightarrow {{\text{T}}_1}{\text{ + }}{{\text{T}}_2}{\text{ + }}{{\text{T}}_3} + {{\text{T}}_4} + {{\text{T}}_5} = $ $\dfrac{1}{4}\left[ {{{\text{T}}_6}{\text{ + }}{{\text{T}}_7}{\text{ + }}{{\text{T}}_8} + {{\text{T}}_9} + {{\text{T}}_{10}}} \right]$
On substituting the values of (i) and (ii) we get,
$ \Rightarrow 5{\text{a + 10d}} = \dfrac{1}{4}\left[ {{\text{5a + 35d}}} \right]$
On solving, we get,
$ \Rightarrow 20{\text{a + 40d}} = {\text{5a + 35d}}$
On taking same terms one side, we get-
$ \Rightarrow 20{\text{a - 5a + 40d - 35d = 0}}$
On solving the above equation, we get-
 $ \Rightarrow 1{\text{5a + 5d = 0}}$
We can write it as-
$ \Rightarrow {\text{5d = - }}1{\text{5a}}$
Then we get,
$ \Rightarrow {\text{d = - 3a}}$
We know that a=$2$. Then we get,
$ \Rightarrow {\text{d = - 6}}$
Now using the value of a and d to find the ${20^{th}}$ term using the formula-
$ \Rightarrow {{\text{T}}_n} = \left( {{\text{a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right)$ where a is the first term, d is the common difference
We get-
$ \Rightarrow {{\text{T}}_{20}} = \left( {{\text{2 + }}\left( {{\text{20 - 1}}} \right)\left( { - 6} \right)} \right)$
On solving we get,
$ \Rightarrow {{\text{T}}_{20}} = {\text{2 - }}\left( {19 \times 6} \right)$
$ \Rightarrow {{\text{T}}_{20}} = {\text{2 - 114 = - 112}}$
Hence Proved

Note: We can also solve this question by using the formula of the sum n terms for A.P. series which is given by-
$ \Rightarrow {{\text{S}}_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where variables have usual notations.
So according to question we can write-
$ \Rightarrow {{\text{S}}_5}{\text{ = }}\dfrac{1}{4}\left( {{{\text{S}}_{10}}{\text{ - }}{{\text{S}}_5}} \right)$ because we can write the sum of next five terms as-${{\text{S}}_5}{\text{ = }}\left( {{{\text{S}}_{10}}{\text{ - }}{{\text{S}}_5}} \right)$
Then putting a=$2$ we will get,
$ \Rightarrow \dfrac{5}{2}\left[ {2 \times 2 + \left( {5 - 1} \right)d} \right] = \dfrac{1}{4}\left[ {\left\{ {\dfrac{{10}}{2}\left( {2 \times 2 + \left( {10 - 1} \right)d} \right)} \right\} - \dfrac{5}{2}\left\{ {2 \times 2 + \left( {5 - 1} \right)d} \right\}} \right]$
On solving we get,
$ \Rightarrow 80 + 80d = 20 + 7d$
On solving this equation we get,
$
   \Rightarrow 10d = - 60 \\
   \Rightarrow d = - 6 \\
 $
Then use the formula of ${{\text{n}}^{{\text{th}}}}$ term,
$ \Rightarrow {{\text{T}}_n} = \left( {{\text{a + }}\left( {{\text{n - 1}}} \right){\text{d}}} \right)$ where a is the first term, d is the common difference to find the ${20^{th}}$ term.
On putting values we get,
$ \Rightarrow {{\text{T}}_{20}} = \left( {{\text{2 + }}\left( {{\text{20 - 1}}} \right)\left( { - 6} \right)} \right)$
On solving we get,
$ \Rightarrow {{\text{T}}_{20}} = {\text{2 - }}\left( {19 \times 6} \right)$
$ \Rightarrow {{\text{T}}_{20}} = {\text{2 - 114 = - 112}}$