If the sum of the first 40 term of the series 3+4+8+9+13+14+18+19+⋯ is 102(m), then the value of m is
A
25
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B
20
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C
10
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D
5
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Solution
The correct option is B20 Let S=(3+4)+(8+9)+(13+14)+(18+19)+⋯ ⇒S=7+17+27+37+⋯upto20 terms
The terms of S are in A.P.
Here a=7,d=10,n=20 ∴S=202[(2×7)+(19×10)](∵Sn=n2[(2a+(n−1)d)]) =202×204 =20×102 ∴m=20