#SEQUENCE & SERIES(CAT 2018)


Find the sum of the series-
7*11 + 11*15 + 15*19 + 19*23 +........+ 95*99

a)80707
b)80705
c)80703
d)80701

SOLUTION-
7*11 + 11*15 + 15*19 + 19*23 +........+ 95*99

General term = (4n+3)(4n+7)

Sum=Σ(4n+3)(4n+7)
       =>Σ(16n^2 + 28n +12n + 21)
       =>Σ(16n^2 + 40n + 21)
       =>16*(n)(n+1)(2n+1)/6 +40(n)(n+1)/2 +21n ------------(i)

Number of terms = (99-11)/4 + 1 = 23
 Sum=(16*23*24*47)/6 + (20*23*24) +21*23
         = 69184 + 11040 + 483 =80707 ANS

Approach-2

Last digit of first term = 7
Last digit of the second term =5
Last digit of third term = 5
Last digit of fourth term = 7
Last digit of fifth term = 1
Now from the sixth term, the cycle repeats,
Sum of last digits of five terms =7+5+5+7+1=25
Now, we know that there are total 23 terms in the series
5*4+3,i.e, the sum of last digits of first 20 terms
5*4=0
Now the sum of the last three terms,7+5+5=7
Hence, the last digit of require sum is 7, only option (A) satisfies.





Comments

  1. Quantose10 April 2019 at 21:42

    You can also use telescope method in this question.

    ReplyDelete

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