Sum of the first six terms of an arithmetic sequence is $9$. Sum of the first twelve terms is $90$. Find the sum of the thirteenth and the seventeenth terms of this sequence.
Let first term of the arithmetic sequence $ = a$
and common difference $= d$
Sum of $n$ terms of an arithmetic sequence $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right) d \right]$
Sum of first six terms $= S_6 = \dfrac{6}{2} \left[2a + 5d\right] = 6a + 15d$
Given: $\;$ $S_6 = 9$
i.e. $\;\;\;$ $6a + 15d = 9$ $\;\;\; \cdots \; (1)$
Sum of first twelve terms $= S_{12} = \dfrac{12}{2} \left[2a + 11d\right] = 12a + 66d$
Given: $\;$ $S_{12} = 90$
i.e. $\;\;\;$ $12a + 66d = 90$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously gives
$a = \dfrac{-7}{2}$ $\;$ and $\;$ $d = 2$
Now, $\;$ $n^{th}$ $\;$ term of an arithmetic sequence $= t_n = a + \left(n - 1\right) d$
$\therefore \;$ $13^{th}$ $\;$ term of the arithmetic sequence $= t_{13} = a + 12d$
and $\;$ $17^{th}$ $\;$ term of the arithmetic sequence $= t_{17} = a + 16d$
$\therefore \;$ Sum of the thirteenth and the seventeenth terms of the sequence
$= t_{13} + t_{17} = a + 12 d + a + 16 d = 2a + 28d$
Substituting the values of $a$ and $d$ gives
$t_{13} + t_{17} = 2 \times \left(\dfrac{-7}{2}\right) + 28 \times 2 = 49$