A) 4000
B) 4020
C) 4200
D) 4220
Correct Answer: B
Solution :
Given \[n=20;{{S}_{20}}=?\] Series (1) \[\to \] 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59... Series (2) \[\to \] 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71. The common terms between both the series are 11, 31, 51, 71... Above series forms an Arithmetic progression (A.P). Therefore, first term (a) = 11 and common difference (d) = 20 Now, \[{{S}_{n}}=\frac{n}{2}[2a+(n-1)d]\] \[{{S}_{20}}=\frac{20}{2}[2\times 11+(20-1)20]\] \[{{S}_{20}}=[22+19\times 20]\] \[{{S}_{20}}=10[22+19\times 20]\] \[{{S}_{20}}=10\times 402=4020\] \[\therefore \]\[{{S}_{20}}=4020\]
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