Question

In an A.P. sum of first ten terms is \[ - {\text{15}}0\] and the sum of its next ten terms is \[ - {\text{55}}0\]. Find the A.P.

Answer 1

Hint :- Make use of the formula of sum of n terms of an arithmetic progression and proceed with the problem.Arithmetic progression is a series of numbers that increases or decreases by the same amount each time.


Complete step by step by solution
We know that sum of n terms.
of an Arithmetic Progression is
\[{{\text{S}}_{\text{n}}}{\text{ }} = {\text{ }}\dfrac{n}{2}{\text{ }}[{\text{2a}} + \left( {{\text{n}} - {\text{1}}} \right)d]\]
Where \[{{\text{S}}_{\text{n}}}\]= Sum of n terms
n = no. of terms
a = first term
d = common difference (${T_{n + 1}} - {T_n}$)
We have given that sum of first ten terms is \[ - {\text{15}}0\]
So,
\[{{\text{S}}_{{\text{10}}}}{\text{ }} = {\text{ }}\dfrac{{10}}{2}{\text{ }}[{\text{2a}} + \left( {10 - {\text{1}}} \right)d] = - 150\]
\[ \Rightarrow {\text{ }}5{\text{ }}[{\text{2a}} + 9d] = - 150\]
\[ \Rightarrow {\text{ 2a}} + 9d = - 30...................(1)\]
Similarly sum of next ten terms is \[ - {\text{55}}0\]
i.e.
${S_{20}} - {S_{10}} = - 550$
${S_{20}} - ( - 150) = - 550$
${S_{20}} = - 700$
\[{\text{ }}\dfrac{{20}}{2}{\text{ }}[{\text{2a}} + \left( {20 - {\text{1}}} \right)d] = - 700\]
\[ \Rightarrow {\text{ }}10{\text{ }}[{\text{2a}} + 19d] = - 700\]
\[ \Rightarrow {\text{ 2a}} + 19d = - 70...................(2)\]
Subtract equation (1) from equation (2)
$ \Rightarrow (2a + 19d) - (2a + 9d) = - 70 - ( - 30)$
$ \Rightarrow 10d = - 40$
$ \Rightarrow d = - 4$
Put the value of d in equation (1)
\[{\text{2a}} + {\text{9(}} - {\text{4}}) = {\text{ }} - {\text{3}}0{\text{ }}\]
\[ \Rightarrow {\text{2a}} - 36 = {\text{ }} - {\text{3}}0{\text{ }}\]
\[ \Rightarrow {\text{2a}} = {\text{ 6 }}\]
\[ \Rightarrow {\text{a}} = {\text{ 3 }}\]
Therefore first term of this A.P. is 3 and the common difference is -4,
Hence A.P. is 3,-1,-5, -9,......


Note –-If the given sequence was a GP, then the sum of the series is given by
Sum of n terms of G.P is $\dfrac{{a({r^n} - 1)}}{{r - 1}}$
Where a is first term,
r is common multiplier,
n is no. of terms.