12. Find the sum of the first 40 positive integers divisible by 6,13.

1.	12. Find the sum of the first 40 positive integers divisible by 6,13. Find the sum of the tirst 15 multiples ofhotuternad 50
Answer» 12. the first 40 positive integers divisible by 6 are 6,12,18,....... upto40 terms the given series is in arthimetic progression with first term a=6 and common difference d=6 sum of n terms of an A.p is n/2×{2a+(n-1)d} →required sum = 40/2×{2(6)+(39)6} =20{12+234} =20×246 =4920 13. The first 8 multiples of 8 are8, 16, 24, 32, 40, 48, 56,64 These are in an A.P., having first term as 8 and common difference as 8. Therefore,a= 8d= 8 S15= ? Sn=n/2[2a+ (n- 1)d] S15= 15/2[2(8)+ (15 - 1)8] =15/2[6 + (14) (8)] =15/2[16 + 112] = 15(128)/2 = 15 × 64 = 960 Like my answer if you find it useful!

12. Find the sum of the first 40 positive integers divisible by 6,13. Find the sum of the tirst 15 multiples ofhotuternad 50

Answer»

12. the first 40 positive integers divisible by 6 are 6,12,18,....... upto40 terms

the given series is in arthimetic progression with first term a=6 and common difference d=6

sum of n terms of an A.p is

n/2×{2a+(n-1)d}

→required sum = 40/2×{2(6)+(39)6}

=20{12+234}

=20×246

=4920

13. The first 8 multiples of 8 are8, 16, 24, 32, 40, 48, 56,64

These are in an A.P., having first term as 8 and common difference as 8.

Therefore,a= 8d= 8

S15= ?

Sn=n/2[2a+ (n- 1)d]

S15= 15/2[2(8)+ (15 - 1)8]

=15/2[6 + (14) (8)]

=15/2[16 + 112]

= 15(128)/2

= 15 × 64

= 960

Like my answer if you find it useful!