If a_1, a_2, a_3, dots are in A.P. such that | vedclass

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(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The terms are given by $a_n = a + (n-1)d$.Given the equation: $a_1 + a_7

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$200$

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(A) Let the first term of the $A.P.$ be $a$ and the common difference be $d$.The terms are given by $a_n = a + (n-1)d$.Given the equation: $a_1 + a_7 + a_{16} = 40$.Substituting the general form: $a + (a + 6d) + (a + 15d) = 40$.This simplifies to: $3a + 21d = 40$.Dividing by $3$: $a + 7d = \frac{40}{3}$.The sum of the first $15$ terms $(S_{15})$ is given by the formula $S_n = \frac{n}{2}[2a + (n-1)d]$.For $n = 15$: $S_{15} = \frac{15}{2}[2a + 14d]$.Factoring out $2$: $S_{15} = 15(a + 7d)$.Substituting the value of $(a + 7d)$: $S_{15} = 15 	imes \frac{40}{3} = 5 	imes 40 = 200$.

If $a_1, a_2, a_3, \dots$ are in $A.P.$ such that $a_1 + a_7 + a_{16} = 40$, then the sum of the first $15$ terms of this $A.P.$ is

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