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)200Option 2)280Option 3)120Option 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

Exams
Articles
Questions