Question

9. Find the sum of first 20 terms of the arithmetic series in which 3rd term is 7 and 7th term
is 2 more than three times its 3rd term.

Answer 1

General term or nth term of A.P
The general term or nth term of A.P is given by an or tn = a + (n – 1)d, where a = a1 is the first term, d is the common difference and n is the number of term.
Sum of n terms of an AP
The sum of first n terms of an AP with first term 'a' and common difference 'd' is given by
Sn = n /2 [ 2a + ( n - 1) d] or
Sn=n /2 [ a + l ] (l = last term)

SOLUTION :

GIVEN :
3rd term = 7
t₃ = 7…………………(1)
t₇ = 3(t₃) + 2
t₇ = 3(7) + 2
[From eq 1]
t₇ = 21 + 2
t₇ = 23

an or tn = a + (n – 1)d
a + 2d = 7 …………….(2)               [t₃ = 7]
a + 6d = 23 ……………(3)              [t₇ = 23]

On Subtracting eq (1) and (2)
a + 2d = 7
a + 6d = 23
(-) (-) (-)
----------------
- 4 d = -16
d = 16/4
d = 4

Put d = 4 in equation 2
a + 2d = 7
a + 2 (4) = 7
a + 8 = 7
a = 7 - 8
a = -1

Sn = (n/2) [ 2a + (n-1) d ]
S₂₀ = (20/2) [ 2(-1) + (20-1) (4) ]
S₂₀ = 10 [ - 2 + 19 (4) ]
S₂₀ = 10 [ - 2 + 76 ]
S₂₀ = 10 [ 74 ]
S₂₀ = 740

Hence, the sum of first 20 terms is 740.

HOPE THIS WILL HELP YOU...

Answer 2

3rd term = 7
a + 2d = 7
a = 7 - 2d ----1equation


7th term = 2 + 3(7) = 2 + 21 = 23
a + 6d = 23
a = 23 - 6d ----1equation


From both equations, we got

7 - 2d = 23 - 6d

7 - 23 = -6d + 2d

- 16 = -4d

$\frac{16}{4} = d$


4 = d



Putting the value of d in 1equation,

a = 7 - 2d
a = 7 - 2(4)
a = 7 - 8
a = -1





Hence, first term = a = -1
Last term = l = 20th term


20th term = a+ 19d
20th term = -1 + 19(4)
20th term = -1 + 76
20th term = 75






$s_{n} = \frac{n}{2} (a + l) \\ \\ => s_{n} = \frac{20}{2} ( - 1 + 75) \\ \\ \\ => s_{n} = 10(74) = 740$


Required sum is 740