\begin{tabular}{|l|l|l|} \hline 29. & $0.7 \times 0.8$ & \\ \hline 30. & $0.8 \times 0.9$ & \\ \hline 31. & $0.8 \times 1.1$ & \\ \hline 32. & $0.8 \times 1.2$ & \\ \hline 33. & $0.8 \times 1.5$ & \\ \hline 34. & $0.9 \times 0.8$ & \\ \hline \end{tabular}

S10 = -150; the sum of next 10 terms = S20 - S10 => -550 = S20 + 150 => S20 = -700; S10 = -150 => 2a1 + 9r = -30; S20 = -700 => 2a1 + 19r = -70; Then 10r = -40 => r = -4 , and, 2a1 -36 = -30 => 2a1 = 6 => a1 = 3;

Answered by crisforp | 2024-06-10

The first term of the arithmetic progression (A.P.) is 3 and the common difference is -4, leading to the sequence 3, -1, -5, and so on. The formula for the nth term of this A.P. is a_n = 3 - 4(n-1). These values satisfy the given conditions about the sums of the terms.
;

Answered by crisforp | 2024-12-20