Question

#Quality Question

@Sequence And Series

Q]]If a1 , a2 , a3 , ...., an are in AP and
a1 + a4 + a7 + ... + a16 = 114.

Then value of a1 + a6 + a11 + a16 is​

Answer 1

$\green{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \:KEY \: \: POINT\: \: \maltese}}}}}$

$\qquad \qquad \mathfrak{Use \: \: nth \: \: term \: \: of \: \: an \: \: \text{ AP} \: \: i.e. } \\ \mathfrak{a_n=a+(n-1) \text{d}} \\\mathfrak{Simplif \text{y }\: \: the \: \: given \: \: equation } \\ \mathfrak{and \: \: use \: \: Result}$

$\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \bigstar \: \: SOLUTION}}}}$

$\mathcal{Given \: A.P. \: is \: a_1,a_2,a_3,...,a_n} \\ \mathcal{ Let \: the \: Above \: AP \: have }\\ \mathcal{ common \: difference \: \bold{d}}$

$\huge \mathcal{Also}$

$\bf a_1 + a_4 + a_7 + ... + a_{16} = 114. \\ \\ \implies \sf a_1 + (a_1 + 3d) + \sf (a_1 + 6d) \\ + \: . \: . \: . \: + \sf (a_1 + 15d) = 114 \\ \\ \implies \sf 6a_1 + 45d = 114 \\ \\ \implies \boxed{ \boxed{\bf2a_1 + 15d = 38}}$

$\huge \mathcal{Now,}$

$\sf a_1 + a_6 + a_{11} + a_{16} \\ \\ \small = \sf a_1 + (a_1 + 5d) + (a_1 + 10d) + (a_1 + 15d) \\ \\ = \sf4a_1 + 30d = 2(2a_1 + 15d) \\ \\ \bf \large = 2 \times 38 = 76$

Answer 2

Given :-

If  a1 , a2 , a3 , ...., an are in AP and

a1 + a4 + a7 + ... + a16 = 114.

To Find :-

Value of a1 + a6 + a11 + a16 is​

Solution :-

So,

The terms are going like

a₁ + a₄ + a₇ + a₁₀ + a₁₃ + a₁₆ = 114

a have repeated 6 times

6/2(a₁ + a₁₆) = 114

3(a₁ + a₁₆) = 114

a₁ + a₁₆ = 114/3

a₁ + a₁₆ = 38 (1)

Now

Now

a₁ + a₆ + a₁₁ + a₁₆

Terms = 4

4/2(a₁ + a₁₆)

4/2 × 38

152/2

76

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