If in an A.P., 3rd term is 18 and 7th term is 30, the sum of its 17 terms is :

Question

If in an A.P., 3rd term is 18 and 7th term is 30, the sum of its 17 terms is : (A) 600 (B) 612 (C) 624 (D) None of these.

Tardigrade Admin · Accepted Answer

T3 = 18 ⇒ a+2d=18 ...(1) T7 = 30 ⇒ a+6d = 30 ...(2) (2) - (1) gives 4d = 12 ⇒ d=3 S17 = (17/2)[2a+(17-1)d] = (17/2)[2a+16d] = 17(a+8d)= 17(a+6d+2d) = 17(30+6) = 17×36 = 612

Q. If in an A.P., 3rd term is 18 and 7th term is 30, the sum of its 17 terms is :

600

612

624

None of these.

$T_{3} = 18$
$\Rightarrow a + 2 d = 18 ... (1)$
$T_{7} = 30$
$\Rightarrow a + 6 d = 30 ... (2)$
$(2) - (1)$ gives $4 d = 12$
$\Rightarrow d = 3$
$S_{17} = \frac{17}{2} [2 a + (17 - 1) d]$
$= \frac{17}{2} [2 a + 16 d]$
$= 17 (a + 8 d) = 17 (a + 6 d + 2 d)$
$= 17 (30 + 6)$
$= 17 \times 36$
$= 612$

Q. If in an A.P., 3rd term is 18 and 7th term is 30, the sum of its 17 terms is :

600

612

624

None of these.

T3​=18 ⇒a+2d=18...(1) T7​=30 ⇒a+6d=30...(2) (2)−(1) gives 4d=12 ⇒d=3 S17​=217​[2a+(17−1)d] =217​[2a+16d] =17(a+8d)=17(a+6d+2d) =17(30+6) =17×36 =612

$T_{3} = 18$
$\Rightarrow a + 2 d = 18 ... (1)$
$T_{7} = 30$
$\Rightarrow a + 6 d = 30 ... (2)$
$(2) - (1)$ gives $4 d = 12$
$\Rightarrow d = 3$
$S_{17} = \frac{17}{2} [2 a + (17 - 1) d]$
$= \frac{17}{2} [2 a + 16 d]$
$= 17 (a + 8 d) = 17 (a + 6 d + 2 d)$
$= 17 (30 + 6)$
$= 17 \times 36$
$= 612$