Question

Question 27
Find the sum of first 17 terms of an AP whose $4^{th}$ and $9^{th}$ terms are – 15 and –30, respectively.

Answer 1

Let the first term, common difference and the number of terms in the AP are a,d and n, respectively.
We know that, the nth term of an AP, $T_n$ = a + (n - 1)d . . . (i)
$\therefore$ $4^{th}$ term of the AP, $T_4$ = a + (4 - 1)d = - 15 [given]
$\Rightarrow$ a + 3d = -15 . . . . . (ii)
$9^{th}$ term of the AP, $T_9$ = a + (9 - 1)d = - 30
[given]
$\Rightarrow$ a + 8d = - 30 ….(iii)
Now, subtract, eq.(ii) from eq.(iii), we get;
$a+8d=-30a+3d=-15--+\underset{–}{}5d=-15d=-3$
$\Rightarrow$
Put the value of d in eq.(ii), we get;
a + 3 (-3) = - 15 $\Rightarrow$ a - 9 = - 15
$\Rightarrow$ a = - 15 + 9 $\Rightarrow$ a = - 6
\Sum of first n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\therefore$ Sum of first 17 terms of the AP, $S_{17}=\frac{17}{2}[2\times (-6)+(17-1\left)\right(-3\left)\right]$
$=\frac{17}{2}[-12+(16\left)\right(-3\left)\right]$
$=\frac{17}{2}(-12-48)=\frac{17}{2}\times (-60)$
$=17\times (-30)=-510$