Question

Tex
(26) Find the number of terms common to the two A.P.
3,7,11,______,407
and 2,9,16,_______,709​

Answer 1

Step-by-step explanation:

Answer in above attachment or picture

Answer 2

AnswEr:

Let the numbers of terms in two A.P.'s be n respectively. Then,

407 = mth term of first A.P. and, 709 = 2 + (n-1) × 7

$\implies \tt407 = 3 + (m - 1) \times 4 \: and \\ \tt709 = 2 + (n - 1) \times 7 \\ \\ \tt \implies \: m = 102 \: \: \: and \: \: \: n = 102.$

So, each A.P. consists of 102 term.

• Let pth term of first A.P. be identical to qth term of the second A.P. then,

$\tt \qquad \: 3 + (p - 1) \times 4 = 2 + (q - 1) \times 7 \\ \\ \rightarrow \tt \: 4p - 1 = 7q - 5 \\ \\ \rightarrow \tt \: 4p + 4 = 7q \\ \\ \rightarrow \tt \: 4(p + 1) = 7q \\ \\ \tt \rightarrow \frac{p + 1}{7} = \frac{q}{4} = k \: \: (say) \\ \\ \rightarrow \tt \: p = 7k - 1 \: \: and \: \: q = 4k$

• Since each A.P. consists of 102 terms.

$\therefore \tt \: p \leqslant 102 \: \: \: \: and \: \: \: \: q \leqslant 102 \\ \\ \implies \tt \: 7k - 1 \leqslant 102 \: \: and \: \: 4k \leqslant 102 \\ \\ \tt \implies \: k \leqslant 14 \frac{5}{7} \: \: and \: \: k \leqslant 25 \frac{1}{2} \\ \\ \implies \tt \: k \leqslant 14 \rightarrow \: k = 1,2,3....,14$

Corresponding to each value of k, we get a pair of identical terms.

• Hence, there are 14 identical terms in two A.P.'s