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find the value of $a 2 a 4 a 6 \dots a {98}$ if $a 1, a 2, a 3, \dots$ is an arithmetic progression with common difference 1, and $a 1 a 2 a 3 \dots a {98}

Respuesta :

value of a2+a4+a6+……+a98 IS EQUAL TO 93.

Given, the common difference (d) =1

Let a1=a Given, a1+a2+……+a98=137

⇒(98/2)(2a+97)=137    [Sn=n/2[2a+(n−1)d]

⇒2a+97=137/49…eq.(1)

a2+a4+…+a98    (n=49 terms)

a2+a4+…+a98=49/2(a2+a98)

a2+a4+…+a98=49/2[(a+1)+a+97]

a2+a4+…+a98=49/2[2a+97+1]

a2+a4+…+a98=49/2[(137/49)+1] [using eq. 1]

a2+a4+…+a98=137/2+49/2

a2+a4+…+a98=186/2 =93

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