Question

Find the sum from the 6th term to the 12th term of the arithmetic progression 6,10,14,...?

Answer 1

a1=6. a=10
d=10

sum from 6th to 12th can be find out by this formula

=> S12-S6
=>
$\frac{12}{2} (2 \times 6 + 11 \times 4) - \frac{6}{2} (2 \times 6 + 5 \times 4) \\ = 6(56) - 3(32) \\ = 336 - 96 \\ = 240$
so your answer is 240

Answer 2

$\Huge{\green{\underline{\textsf{Answer}}}}$

$\large{\red{\underline{\tt{Given:}}}}$

$\implies\rm a=6$

$\implies\rm d=4$

$\large{\red{\underline{\tt{Solution:}}}}$

$\implies\rm T6= a+5d =6+20=26$

$\implies\rm T12= a+11d =6+44=50$

$\implies\rm Here,\: from\:T6\:and\:T12\:we\:have\:7\:terms$

$\implies\rm Therefore\: n=7$

$\implies\rm 7(26+50)/2$

$\implies\rm 7×76/2$

$\implies\rm 266$