For an A.P.,the 3^{rd} term is 7 and the 7^{t | vedclass

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(B) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.$T_n = a + (n - 1)d$Given $T_3 = 7$,so $a + 2d = 7$  --- $(1)$Given $T_7

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$740$

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(B) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.$T_n = a + (n - 1)d$Given $T_3 = 7$,so $a + 2d = 7$  --- $(1)$Given $T_7 = 3(T_3) + 2$,so $a + 6d = 3(7) + 2 = 23$  --- $(2)$Subtracting equation $(1)$ from $(2)$:$(a + 6d) - (a + 2d) = 23 - 7$$4d = 16$$d = 4$Substituting $d = 4$ in equation $(1)$:$a + 2(4) = 7$$a + 8 = 7$$a = -1$The sum of the first $n$ terms is given by $S_n = \frac{n}{2}[2a + (n - 1)d]$.For $n = 20$:$S_{20} = \frac{20}{2}[2(-1) + (20 - 1)4]$$S_{20} = 10[-2 + 19(4)]$$S_{20} = 10[-2 + 76]$$S_{20} = 10[74] = 740$.Thus,the sum of the first $20$ terms is $740$.

Column $A$	Column $B$
$(A_{1}) \quad 2, -2, -6, -10, \ldots$	$(B_{1}) \quad \frac{2}{3}$
$(A_{2}) \quad a = -18, n = 10, a_{n} = 0$	$(B_{2}) \quad -5$
$(A_{3}) \quad a = 0, a_{10} = 6$	$(B_{3}) \quad 4$
$(A_{4}) \quad a_{2} = 13, a_{4} = 3$	$(B_{4}) \quad -4$
	$(B_{5}) \quad 2$
	$(B_{6}) \quad \frac{1}{2}$
	$(B_{7}) \quad 5$

For an $A.P.$,the $3^{rd}$ term is $7$ and the $7^{th}$ term exceeds thrice the $3^{rd}$ term by $2$. Find the sum of first $20$ terms of the $A.P.$

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