For an A.P.,the first term is 10,the n^{th} t | vedclass

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(A) Given: First term $a = 10$,$n^{th}$ term $a_n = 40$,and sum of $n$ terms $S_n = 150$.We know the formula for the sum of an $A.P.$ is $S_n = \frac{

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$n=6, d=6$

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(A) Given: First term $a = 10$,$n^{th}$ term $a_n = 40$,and sum of $n$ terms $S_n = 150$.We know the formula for the sum of an $A.P.$ is $S_n = \frac{n}{2}(a + a_n)$.Substituting the given values: $150 = \frac{n}{2}(10 + 40)$.$150 = \frac{n}{2}(50)$.$150 = 25n$.$n = \frac{150}{25} = 6$.Now,we use the formula for the $n^{th}$ term: $a_n = a + (n - 1)d$.Substituting the values: $40 = 10 + (6 - 1)d$.$40 - 10 = 5d$.$30 = 5d$.$d = \frac{30}{5} = 6$.Thus,$n = 6$ and $d = 6$.

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 first term is $10$,the $n^{th}$ term is $40$,and the sum of the first $n$ terms is $150$. Find the value of $n$ and the common difference $d$.

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