Question

In the sequence 2,5, 8…. Up to 50 terms and 3,5, 7…. Up to 60 terms, find the number of identical terms.

Answer 1

Hint: Since series of identical terms of two AP are always AP in therefore for finding the number of term we will use ${l_B} = A + \left( {n - 1} \right)d$ in which 1st common term is 5 and common difference is LCM of the common difference is of two given series have to find last terms of the AP series first we have to calculate the last term of series 2,5, 8…. A=2, d=3 and n=50 and in series 3,5, 7…. A=3, d=2 and n=60. Then after comparing the last terms, the last term of the required AP is always less than the smallest last term in the given series.

Complete step-by-step answer:
In series 2,5, 8…. A=2, d=3 and n=50 therefore last term will be
${l_A} = A + \left( {n - 1} \right)d$
$\Rightarrow$ ${l_A} = 2 + \left( {50 - 1} \right)3 = 149$
In series 3, 5, 7 …. A=3, d=2 and n=60 therefore last term will be
${l_B} = A + \left( {n - 1} \right)d$
$\Rightarrow$ ${l_B} = 3 + \left( {60 - 1} \right)2 = 121$
Identical terms must be in AP where 5 is the 1st common term and common difference is LCM of the common difference 2, 5, 8 …. and common difference of 3,5, 7….
LCM of 3 and 2 =$3 \times 2 = 6$
Now, last term of the AP of identical term will be less than 121
$A + \left( {n - 1} \right)d \leqslant 121$
Substituting A=5 and d=6
$\Rightarrow$ $5 + \left( {n - 1} \right)6 \leqslant 121$
$\Rightarrow$ $\left( {n - 1} \right) \leqslant \dfrac{{121 - 5}}{6}$
$\Rightarrow$ $\left( {n - 1} \right) \leqslant 19.3333$
n is integer therefore n-1 cannot be a fraction it will be a integer
$\therefore n - 1 = 19$
So, $n = 20$
No. of identical terms is 20.

Note: Point to remember series of identical terms of two AP are always AP. And the common difference is LCM of the common difference is of given series. n is always integers it will not be a fraction therefore we select the nearest smallest integer.