Question

In an arithmetic series, the sum of the first 14 terms is – 203 and the sum of the next 11 terms is –572. Find the arithmetic series.

Answer 1

Hint: - Assume that the first term and common difference of the arithmetic series is ‘a’ and ‘d’ respectively. Use the formula for sum of n terms of the arithmetic series that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, Now form two equations using this formula and solve to get the value of a and d.

Complete step-by-step answer:
Let the first term of the arithmetic series will be a.
And the common difference of the arithmetic series is d.
So, now as it is given that the sum of first 14 term of this series is equal to –203
And the formula for sum of first n terms of the arithmetic series is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$
So, ${S_{14}} = \dfrac{{14}}{2}\left[ {2a + \left( {14 - 1} \right)d} \right] = - 203$
$\Rightarrow 2a + 13d = - 29$ (1)
Now the sum of the next 11 terms is equal to –572. i.e. in other words we can say that the sum of all terms between 15th term and 25th term including them is equal to –572.
Now as all the terms are in arithmetic series. So, all terms between the 15th and 25th term will also form an arithmetic series.
Now as we know that there is another formula for the sum of n terms of the series.
${S_n} = \dfrac{n}{2}\left[ {{T_{First}} + {T_{Last}}} \right]$ where ${T_{First}}$ will be the first term from which we had to find the sum (i.e. 15th term) and ${T_{Last}}$ will be the last term till which we had to find the sum (i.e. 25th term)
So, to find the sum of the next 11 terms we had to first find the 15th and 25th term of the series.
Now according to the formula of arithmetic series nth term of the series is calculated as ${T_n} = a + \left( {n - 1} \right)d$
So, ${T_{15}} = a + \left( {15 - 1} \right)d = a + 14d$
And, ${T_{25}} = a + \left( {25 - 1} \right)d = a + 24d$
Now finding the sum of the next 11 terms of the arithmetic series.
$\Rightarrow {S_{11}} = \dfrac{{11}}{2}\left[ {{T_{15}} + {T_{25}}} \right] = \dfrac{{11}}{2}\left[ {a + 14d + a + 24d} \right] = \dfrac{{11}}{2}\left[ {2a + 38d} \right] = 11\left( {a + 19d} \right)$
$\Rightarrow a + 19d = - 52$ (2)
Now we had to solve equation 1 and 2 to find the value of a and d.
So, multiplying equation 2 by 2 and then subtracting from equation 1 from equation 2.

\Rightarrow 2a + 38d - 2a - 13d = - 104 + 29 \\\ \Rightarrow 25d = - 75 \\\ \end{gathered} $$ So, d = –3 and on putting the value of d in equation 2. We get, a + 19*(–3) = –52 a – 57 = –52 a = 57 – 52 = 5 So, now first term = a = 5 and common difference = d = –3 So, the A.P will be 5, (5 +(–3)), (5 + 2(–3)), (5 + 3(–3)), (5 + 4(–3)), (5 + 5(–3))……. Hence, arithmetic series will be 5, 2, –1, –4, –7, –10……. Note:- Whenever we come up with this type of problem then the first, we have to assume the first term and common difference as a and d. And after that we form equations using the formula of sum of n terms $${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] = \dfrac{n}{2}\left[ {{T_{First}} + {T_{Last}}} \right]$$ and nth term of the series $${T_n} = a + \left( {n - 1} \right)d$$. And on solving these equations by substitution method we will get the value of a and d. After getting the value of a and d we can find all the terms of the A.P using formula $${T_n} = a + \left( {n - 1} \right)d$$. This will be the easiest and efficient way to find the solution of the problem.