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

a1+a4+a7+….+a16=114⇒3a1+a16=114⇒a1+a16=38 Now, a1+a6+a11+a16=2a1+a16=2×38=76

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

a1+a4+a7+….+a16=114⇒3a1+a16=114⇒a1+a16=38 Now, a1+a6+a11+a16=2a1+a16=2×38=76

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

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$a_{1} + a_{4} + a_{7} + \dots . + a_{16} = 114$

$\Rightarrow 3 (a_{1} + a_{16}) = 114$

$\Rightarrow a_{1} + a_{16} = 38$

$Now, a_{1} + a_{6} + a_{11} + a_{16} = 2 (a_{1} + a_{16}) = 2 \times 38 = 76$

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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

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