Question

Find the number of terms common to the two A,P's $3,7,11,.....,407$ and $2,9,16,..........,709$

• A. $12$
• B. $14$
• C. $25$
• D. $28$

Answer 1

The correct option is C $25$

The given terms of $AP$ are $3,7,\mathrm{11.........}407$ and $2,916,......70$

Now,

Let $m^{th}$ term at first sequence be same as the $n^{th}$ terms at the second sequence Then $t_m$ and $t_n$

$\because 3+\left(m-1\right)\times 4=2+\left(n-1\right)7$

$\Rightarrow 4m-1=n-5$

$\Rightarrow 4m-7n=-5+1$

$\Rightarrow 4m=7n-4$

$\Rightarrow m=\frac{7n-4}{4}$

Now, since n is a whole number, so 7m-4 needs to be divisible by 4. So, then, m can be equal to all multiples of 4 till 102 i.e. 4, 8, 12, 16, 20, 24, ......100So, the number of terms common to the 2 APs would be 25.

Hence, the option $\left(C\right)$ is correct answer.