Question

FIND THE SUM OF 20 TERMS OF THE ARITHMETIC SERIES IN WHICH 3rd TERM IS 7 AND 7th TERM IS 2 MORE THAN THREE TIMES ITS 3rdTERM

Answer 1

Answer: 740


Let the AP be a, a+d, a+2d, a+3d, ...


The $n^{th}$ term of AP is given by:

$T_n = a+(n-1)d$

We are given that the third term is 7.

$T_3=7 \\ \\ \implies a+(3-1)d=7 \\ \\ \implies a+2d=7 \quad ---(1)$


Also, the seventh term is 2 more than three times the third term.

$T_7 = 3T_3+2 \\ \\ \implies a+(7-1)d = 3(a+(3-1)d)+2 \\ \\ \implies a+6d=3(a+2d)+2 \\ \\ \implies a+6d = 3a+6d+2 \\ \\ \implies -2a=2 \\ \\ \implies \bold{a=-1}$

We can put it in (1)

$a+2d=7\\ \\ \implies (-1)+2d=7 \\ \\ \implies 2d=8 \\ \\ \implies \bold{d=4}$



Now, the Sum of n terms of an AP is given by the formula:

$S_n = \frac{n}{2}(2a+(n-1)d)$

So, Sum of first 20 terms would be:


$S_{20} = \frac{20}{2} (2(-1)+(20-1)\times 4) \\ \\ \implies S_{20}=10(-2+19\times 4) \\ \\ \implies S_{20} = 10 \times (-2+76) \\ \\ \implies S_{20} = 10 \times 74 \\ \\ \implies \boxed{\bold{S_{20}=740}}$


Thus, The Sum of first 20 terms of the AP is 740.

Answer 2

Answer:

740

Step-by-step explanation:

s20=?

a+2d=7

a+6d=16

a-a+6d-2d=23-7

4d=16

d=4

s20=n/2(2a+(n-1)d

=10(-2+(19)(4))

=74*10

=740