#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, w e 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.
