Let a_1, a_2, a_3, ldots be an arithmetic pro | vedclass

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(B) Given $a_1=7$ and $d=8$. The $n$-th term of the arithmetic progression is $a_n = a_1 + (n-1)d = 7 + (n-1)8 = 8n-1$.We are given $T_{n+1} - T_n = a

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$B, C$

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(B) Given $a_1=7$ and $d=8$. The $n$-th term of the arithmetic progression is $a_n = a_1 + (n-1)d = 7 + (n-1)8 = 8n-1$.We are given $T_{n+1} - T_n = a_n$. Summing this from $k=1$ to $n-1$,we get $T_n = T_1 + \sum_{k=1}^{n-1} a_k$.$T_n = 3 + \sum_{k=1}^{n-1} (8k-1) = 3 + 8 \frac{(n-1)n}{2} - (n-1) = 3 + 4n^2 - 4n - n + 1 = 4n^2 - 5n + 4$.For $n=20$,$T_{20} = 4(20)^2 - 5(20) + 4 = 1600 - 100 + 4 = 1504$. (Option $A$ is false).For $n=30$,$T_{30} = 4(30)^2 - 5(30) + 4 = 3600 - 150 + 4 = 3454$. (Option $C$ is true).Now,$\sum_{k=1}^n T_k = \sum_{k=1}^n (4k^2 - 5k + 4) = 4 \frac{n(n+1)(2n+1)}{6} - 5 \frac{n(n+1)}{2} + 4n$.For $n=20$,$\sum_{k=1}^{20} T_k = 4 \frac{20(21)(41)}{6} - 5 \frac{20(21)}{2} + 4(20) = 2(2870) - 1050 + 80 = 11480 - 1050 + 80 = 10510$. (Option $B$ is true).For $n=30$,$\sum_{k=1}^{30} T_k = 4 \frac{30(31)(61)}{6} - 5 \frac{30(31)}{2} + 4(30) = 2(18920) - 2325 + 120 = 37840 - 2325 + 120 = 35635$. (Option $D$ is false).Thus,options $B$ and $C$ are true.

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