If a_1,a_2,a_3,................. 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 :

  • Option 1)

    200

  • Option 2)

    280

  • Option 3)

    120

  • Option 4)

    150

Answers (1)

Given , a_1+a_7+a_{16}=40

       =>a_1+(a_1+6d)+(a_1+15d)=40

       =>a_1+7d=\frac{40}{3}..................(1)

We have to find out ,

a_1,a_2,a_3,.................,a_{15}

=> \frac{15}{2}[a_1+a_{15}]=\frac{15}{2}[a_1+a_1+14d]=15(a_1+7d)..............(2)

Substituting the value of (1) in (2),

=>15\times \frac{40}{3}=200


Option 1)

200

Option 2)

280

Option 3)

120

Option 4)

150

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