Question

The 5th term of an AP is 23 and 12th term is 37. Find the first term and common difference

Answer 1

Answer :-

First term = 15

Common difference = 2

Solution :-

Fifth term ($\tt{T_{5}}$) = 23

Twelfth term ($\tt{T_{12}}$) = 37

We know that,

$\tt{T_{n} = a + (n - 1)d}$

where,

a = First term of an A.P.,

n = Number of terms of an A.P.,

d = Difference of consecutive terms of an A.P.

So,

=> $\tt{T_{5}}$ = a + (5 - 1)d

=> 23 = a + (4)d

=> 23 = a + 4d

=> 23 - 4d = a ... eq. i.)

Now,

=> $\tt{T_{12}}$ = a + (12 - 1)d

=> 37 = a + (11)d

=> 37 = a + 11d

Substituting the value of a from eq. i.)

=> 37 = (23 - 4d) + 11d

=> 37 = 23 - 4d + 11d

=> 37 = 23 + 7d

=> 37 - 23 = 7d

=> 14 = 7d

=> $\tt{\dfrac{14}{7}=d}$

=> 2 = d

Substituting the value of d in eq. i.)

=> 23 - 4d = a

=> 23 - 4 × 2 = a

=> 23 - 8 = a

=> 15 = a

First term = a = 15

Common difference = d = 2