JEE Mains · Maths · STD 11 - 8. sequence and series
The common difference of the \(A.P.:a_{1},a_{2},....,a_{m}\) is 13 more than the common difference of the \(A.P.: b_{1}, b_{2},...,b_{n}.\) If \(b_{31}=-277, b_{43}=-385\) and \(a_{78}=327,\) then \(a_{1}\) is equal to
- A 21
- B 24
- C 19
- D 16
Answer & Solution
Correct Answer
(C) 19
Step-by-step Solution
Detailed explanation
Let common difference of A.P.'s are \(d_{1}\) & \(d_{2}\) \(\therefore\) \(d_{1}=13+d_{2}\) \(b_{1}+30d_{2}=-277\) ..(1) \(b_{1}+42d_{2}=-385\) .....(2) By (2) - (1) \(12 \mathrm{~d}_2=-108\) \(\mathrm{d}_2=-9\) \(\therefore \mathrm{d}_1=4\) Now \(\mathrm{a}_{78}=327\)…
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