Question

In an ap the sum of first ten terms is minus 150 and the sum of its next 10 term is minus 550 find the ap

Answer 1

$d = common\ difference \\ \\ n = no.\ of\ terms \\ \\ a_1 = first\ term \\ \\ a_2 = second\ term \\ \\ a_3 = third\ term$

$\\ \\ \\ We\ know\ that\ the\ difference\ between\ the\ sums\ of\ first\ n\ terms \\ and\ next\ n\ terms\ is\ n^2d. \\ \\ \therefore\ n^2d = (-550) - (-150) \\ \\ = 10^2d = - 550 + 150 \\ \\ = 100d = -(550 - 150) = - 400 \\ \\ d = \frac{-400}{100} = -4 \\ \\ \\ Let's\ find\ the\ first\ term.$

$\\ \\ \\ Sum\ of\ first\ 10\ terms \\ \\ = \frac{10}{2}[2a_1 + 9d] = -150 \\ \\ = 5[2a_1 + 9 \times (-4)] = -150 \\ \\ = 5[2a_1 - 36] = -150 \\ \\ 2a_1 - 36 = \frac{-150}{5} = -30 \\ \\ 2a_1 = - 30 + 36 = 36 - 30 = 6 \\ \\ a_1 = \frac{6}{2} = 3 \\ \\ \\ a_1 = 3 \\ a_2 = 3 - 4 = -1 \\ a_3 = -1 - 4 = -5 \\ \\ \\ \therefore\ 3,\ -1,\ -5,......\ is\ the\ AP.$

$Thank\ you. \\ \\ \\ \#adithyasajeevan$