Question

JEE MAINS MATHS QUESTION SEPTEMBER 2020

Answer 1

ANSWER.

=> value of B1 = -81

=> option [ c ] is correct option.

EXPLANATION.

$\sf \to \: a_{1}, a_{2}, a_{3},....... a_{n} \: \: are \: in \: ap \\ \\ \sf \to \: b_{1}, b_{2}, b_{3},.......b_{n} \: \: \: are \: in \: ap \\ \\ \sf \to \: common \: difference \: of \: second \: is \: more \: than \: the \: first \: \\ \\ \sf \to \: D \: = d + 2$

$\sf \to \: a_{40 } = - 159 \\ \\ \sf \to \: a + 39d = - 159.....(1) \\ \\ \sf \to \: a_{100} = - 399 \\ \\ \sf \to \: a \: + 99d = - 399.....(2) \\ \\ \sf \to \: from \: equation \: (1) \: \: and \: \: (2) \: \: we \: get$

$\sf \to \: - 60d \: = 240 \\ \\ \sf \to \: d \: = - 4 \\ \\ \sf \to \: put \: the \: value \: of \: d \: = - 4 \: in \: equation \: (1) \: we \: get \\ \\ \sf \to \: a \: + 39( - 4) = - 159 \\ \\ \sf \to \: a - 156 = - 159 \\ \\ \sf \to \: a \: = - 3$

$\sf \to \: b_{100} = a_{70} \\ \\ \sf \to \: b_{1} \: + 99D \: = a + 69d \: \\ \\ \sf \to \: b_{1} + 99 \times ( - 2) = - 3 + 69 \times ( - 4) \\ \\ \sf \to \: b_{1} + (- 198 ) = - 3 - 276 \\ \\ \sf \to \: b_{1} = - 81$