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**The 3rd and 8th term of an arithmetic progression are -13 and 2 respectively. What is the 14th term?**

A. 23
B. 17
C. 20
D. 26
**Answer: Option C**

## Show Answer

Solution(By Apex Team)

Let the first term of an AP = a and the common difference = d
3rd term of AP = A3 = a + 2d = -13 …… (1)
8th term = A8 = a + 7d = 2 …… (2)
Subtracting equation (1) from (2), we get :
⇒ 7d – 2d = 2 – (-13)
⇒ 5d = 15
⇒ d = $\Large\frac{15}{5}$ = 3
Substituting it in equation (2)
⇒ a = 2 – 7(3) = 2 – 21 = -19
∴ 14th term = A14 = a + 13d
= -19 + 13(3)
= -19 + 39
= 20

## 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