If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 1 3 + 14 + 18 + 19 + ... is (102)m, then m is equal to :

Question

If the sum of the first 40 terms of the series, 3 + 4 + 8 + 9 + 1 3 + 14 + 18 + 19 + ... is (102)m, then m is equal to : (A) 10 (B) 25 (C) 5 (D) 20

Tardigrade Admin · Accepted Answer

Sum of the 40 terms of 3 + 4 + 8 + 9 + 13 + 14 + 18 + 19… = (3 + 8 + 13 +…upto 20 term) + [4 + 9 + 15 +… upto 20 terms] =10[ 6 + 19 × 5 + 8 + 19 × 5 ] = 10 × 204 = 20 × 102

Q. If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ...$ is $(102) m$ , then m is equal to :

$10$

$25$

$5$

$20$

Sum of the 40 terms of
$3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 \dots$
$= (3 + 8 + 13 + \dots u pt o 20 t er m)$
$+ [4 + 9 + 15 + \dots u pt o 20 t er m s]$
$= 10 [{6 + 19 \times 5 + 8 + 19 \times 5}]$
$= 10 \times 204 = 20 \times 102$

Q. If the sum of the first 40 terms of the series, 3+4+8+9+13+14+18+19+... is (102)m, then m is equal to :

10

25

5

20

Sum of the 40 terms of 3+4+8+9+13+14+18+19… =(3+8+13+…upto20term) +[4+9+15+…upto20terms] =10[{6+19×5+8+19×5}] =10×204=20×102

Q. If the sum of the first 40 terms of the series, $3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + ...$ is $(102) m$ , then m is equal to :

$10$

$25$

$5$

$20$

Sum of the 40 terms of
$3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 \dots$
$= (3 + 8 + 13 + \dots u pt o 20 t er m)$
$+ [4 + 9 + 15 + \dots u pt o 20 t er m s]$
$= 10 [{6 + 19 \times 5 + 8 + 19 \times 5}]$
$= 10 \times 204 = 20 \times 102$