Question

If the first term, second term, and the last term of an A.P are 6, 9,and 33 respectively, then find the number of terms

Answer 1

Answer:

n= 10

Step-by-step explanation:

a1 = 6

a2 = 9

d = a1-a2

 = 3

an = a+(n-1)d

33 = 6 + (n-1) 3

27= 3n-3

3n = 30

n= 10

Answer 2

Given ,

• $\displaystyle\sf{X_1=6}$

• $\displaystyle\sf{X_2=9}$

• $\displaystyle\sf{X_n=33}$

• $\displaystyle\sf{d=9-6=3}$

• $\displaystyle\sf{A.P=6,9,12,...,33}$

Here, we have to find the number of terms.

The number of terms in an A.P. can be determined by the equation,

$\boxed{\displaystyle\sf{n=(\frac{X_n-X_1}{d})+1}}$

Applying the values to the equation ,

$\implies\displaystyle\sf{n=(\frac{33-6}{3})+1}$

$\implies\displaystyle\sf{n=\frac{27}{3}+1}$

$\implies\displaystyle\sf{n=\frac{3+27}{3}}$

$\implies\displaystyle\sf{n=\frac{30}{3}}$

$\implies\displaystyle\sf{n=10}$

$\boxed{\displaystyle\sf{\therefore Number\:of\:Terms=10}}$