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**The 3rd and 9th term of an arithmetic progression are -8 and 10 respectively. What is the 16th term?**

A. 34
B. 28
C. 25
D. 31
**Answer: Option D**

## Show Answer

Solution(By Apex Team)

Let the first term of an AP = a and the common difference = d
3th term of AP = A3 = a + 2d = -8 ……(1)
9th term = A9 = a + 8d = 10 …… (2)
Subtracting equation (1) from (2), we get :
⇒ 8d – 2d = 10 – (-8)
⇒ 6d =18
⇒ d = $\Large\frac{18}{6}$ = 3
$\begin{array}{l}\text{ Substituting it in equation (2), }\\
\begin{aligned}\Rightarrow a&=10-8(3)\\
&=10-24\\
&=-14\end{aligned}\end{array}$
$\begin{array}{l}\therefore16^{\text{th }}\text{ term }=\mathrm{A}_{16}=\mathrm{a}+15\mathrm{~d}\\
=-14+15(3)\\
=-14+45\\
=31\end{array}$

## Related Questions On Progressions

### How many terms are there in 20, 25, 30 . . . . . . 140?

A. 22B. 25

C. 23

D. 24

### Find the first term of an AP whose 8th and 12th terms are respectively 39 and 59.

A. 5B. 6

C. 4

D. 3

### Find the 15th term of the sequence 20, 15, 10 . . .

A. -45B. -55

C. -50

D. 0

### The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is

A. 600B. 765

C. 640

D. 680