If a1- a2- a3- - an are in A-P- and a1-a4-a7-a16-114- then a1-a6-a11-a16 is equal to -A- 76B- 98C- 38D- 64

The correct option is A 76We now that since, the series is in nbsp; A.P. ∴ a_1+a_16=a_4+a_13=a_7+a_10 =a_5+a_12=a_6+a_11 ⇒ 3(a_6+a_11)=114 ⇒ nbsp; a_6+a_11 ...

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Byju's Answer

Standard XII

Mathematics

Arithmetic Progression

If a1, a2, a3...

Question

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in A.P. and $a_{1} + a_{4} + a_{7} + \dots + a_{16} = 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to :

38

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64

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76

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98

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Solution

The correct option is C $76$
We now that since, the series is in A.P.
$∴ a_{1} + a_{16} = a_{4} + a_{13} = a_{7} + a_{10} = a_{5} + a_{12} = a_{6} + a_{11}$
$\Rightarrow 3 (a_{6} + a_{11}) = 114 \Rightarrow a_{6} + a_{11} = 38 \Rightarrow (a_{1} + a_{6} + a_{11} + a_{16}) = 2 (a_{6} + a_{11}) = 2 \times 38 = 76$

Suggest Corrections

then $a_{1} + a_{6} + a_{11} + a_{16} =$

Q. If for an

A . P .

a_{1}, a_{2}, a_{3}, . . . . . ., a_{n}, . . . . .

a_{1} + a_{3} + a_{5} = - 12

and

a_{1} a_{2} a_{3} = 8

then the value of

a_{2} + a_{4} + a_{6}

equals

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If a1,a2,a3,⋯,an are in A.P. and a1+a4+a7+⋯+a16=114 , then a1+a6+a11+a16 is equal to :

The correct option is C 76We now that since, the series is in A.P. ∴a1+a16=a4+a13=a7+a10=a5+a12=a6+a11 ⇒3(a6+a11)=114⇒a6+a11=38⇒(a1+a6+a11+a16)=2(a6+a11) =2×38 =76

then a1+a6+a11+a16=

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ are in A.P. and $a_{1} + a_{4} + a_{7} + \dots + a_{16} = 114$ , then $a_{1} + a_{6} + a_{11} + a_{16}$ is equal to :

then $a_{1} + a_{6} + a_{11} + a_{16} =$