If a 1- a 2- a 3- - is an arithmetic progression with common difference 1 and a 1- a 2- a 3- a 98-137- then what is the value of a 2- a 4- a 6- a 98 - -2 MARKS-

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Byju's Answer

Standard X

Mathematics

General Form of an AP

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, a_{3}, \dots$ is an arithmetic progression with common difference 1 and $a_{1} + a_{2} + a_{3} + \dots \dots + a_{98} = 137$ , then what is the value of $a_{2} + a_{4} + a_{6} + \dots \dots + a_{98}$ ? [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Given, the common difference (d) =1
$L e t a_{1} = a$

$G i v e n, a_{1} + a_{2} + \dots \dots + a_{98} = 137$

$\Rightarrow \frac{98}{2} [2 a + 97] = 137 [S_{n} = \frac{n}{2} [2 a + (n - 1) d]$

$\Rightarrow 2 a + 97 = \frac{137}{49} \dots (i)$

$a_{2} + a_{4} + \dots + a_{98} (49 t e r m s)$

$= \frac{49}{2} [a_{2} + a_{98}]$

$= \frac{49}{2} [(a + 1) + a + 97]$

$= \frac{49}{2} [2 a + 97 + 1]$

$= \frac{49}{2} [\frac{137}{49} + 1]$ [From (i)]

$= \frac{137}{2} + \frac{49}{2}$

$= \frac{186}{2}$

$= 93$

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If a1,a2,a3, … is an arithmetic progression with common difference 1 and a1+a2+a3+……+a98=137, then what is the value of a2+a4+a6+……+a98? [2 MARKS]

Concept: 1 Mark Application: 1 Mark Given, the common difference (d) =1 Let a1=a Given, a1+a2+……+a98=137 ⇒982[2a+97]=137 [Sn=n2[2a+(n−1)d] ⇒2a+97=13749…(i) a2+a4+…+a98 (49 terms) =492[a2+a98] =492[(a+1)+a+97] =492[2a+97+1] =492[13749+1] [From (i)] =1372+492 =1862 =93

If $a_{1}, a_{2}, a_{3}, \dots$ is an arithmetic progression with common difference 1 and $a_{1} + a_{2} + a_{3} + \dots \dots + a_{98} = 137$ , then what is the value of $a_{2} + a_{4} + a_{6} + \dots \dots + a_{98}$ ? [2 MARKS]