JEE MAIN - Mathematics (2019 - 12th April Evening Slot - No. 8)
If a1, a2, a3, ..... are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is :
120
200
150
280
Explanation
a1 + a7 + a16 = 40
$${a_1} + \left( {{a_1} + 6d} \right) + ({a_1} + 15d) = 40$$
$$ \Rightarrow 3{a_1} + 21d = 40$$
$$ \Rightarrow {a_1} + 7d = {{40} \over 3}$$
$$ \Rightarrow {a_1} + {a_2}....... + {a_{15}} = {{15} \over 2}[{a_1} + {a_{15}}]$$
$$ \Rightarrow {{15} \over 2}[{a_1} + {a_1} + 14d] \Rightarrow 15({a_1} + 7d) = 15 \times {{40} \over 3} = 200$$
$${a_1} + \left( {{a_1} + 6d} \right) + ({a_1} + 15d) = 40$$
$$ \Rightarrow 3{a_1} + 21d = 40$$
$$ \Rightarrow {a_1} + 7d = {{40} \over 3}$$
$$ \Rightarrow {a_1} + {a_2}....... + {a_{15}} = {{15} \over 2}[{a_1} + {a_{15}}]$$
$$ \Rightarrow {{15} \over 2}[{a_1} + {a_1} + 14d] \Rightarrow 15({a_1} + 7d) = 15 \times {{40} \over 3} = 200$$
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