Question ID: #930
The common difference of the A.P.: $a_{1}, a_{2}, \dots, a_{m}$ is 13 more than the common difference of the A.P.: $b_{1}, b_{2}, \dots, b_{n}.$ If $b_{43}=-385$ and $b_{31}=-277$, and $a_{78}=327$, then $a_{1}$ is equal to
- (1) 21
- (2) 24
- (3) 19
- (4) 16
Solution:
Let $d_1$ be the common difference of sequence $a$ and $d_2$ be the common difference of sequence $b$.
Given $d_1 = d_2 + 13$.
For sequence $b$:
$$b_{43} = b_1 + 42d_2 = -385 \quad \dots(1)$$
$$b_{31} = b_1 + 30d_2 = -277 \quad \dots(2)$$
Subtract (2) from (1):
$$12d_2 = -108$$
$$d_2 = -9$$
Now find $d_1$:
$$d_1 = -9 + 13 = 4$$
For sequence $a$, we are given $a_{78} = 327$:
$$a_{78} = a_1 + 77d_1$$
$$327 = a_1 + 77(4)$$
$$327 = a_1 + 308$$
$$a_1 = 327 – 308$$
$$a_1 = 19$$
Ans. (3)
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