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Find the $S_{15}$ if $T_3 = 18$ and $T_{10} = 67$.

Find the sum in this AP.

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    I'm not really sure how to pose a question on here. Bear with me. Thanks – 2017-02-25
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    What have you tried? Can you write the general form of a term of an arithmetic progression? – 2017-02-25

3 Answers 3

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Well if $a$ is the first term and $d$ is the common difference then $a + 2d = 18$ and $a + 9d = 67$, subtracting the two gives $7d = 49 \implies d = 7$ so $a = 4$.

Now the sum of the first $15$ terms is $\frac{15}{2}(4 + 102) = 795 $.

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In general for APs: $T_{n} = T_{n-1} + d, n\geq 2$ and $d$ is the common difference between terms

There it is easy to see that:

$T_{10} = T_{3} + 7d$

$67 = 18 + 7d$

$d = 7$

Now the formula for the sum,$S_{n}$, to $n$ terms of an AP (let me know if you want the proof):

$S_{n} = \frac{n}{2}(2a+(n-1)d), a = T_{1}$

We can find $T_{1} = T_{3} - 2d = 4$

$\Rightarrow S_{n} = \frac{n}{2}(8 +7(n-1))$ = $\frac{n}{2}(7n+1)$

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By linear interpolation, $$T_8=53$$ Hence, using $S_{2n-1}=(2n-1)T_n$, $$S_{15}=15\;T_8=\color{red}{795}$$