In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms?
Answers
Answer:
The sum of first 20 terms is 1150.
Step-by-step explanation:
Given :
a5 = 30, a12 = 65 ,
Case 1 :
By using the formula ,an = a + (n - 1)d
a5 = 30
a + (5 - 1)d = 30
a + 4d = 30
a = 30 – 4d…………... (1)
Case 2:
a12 = 65
a + (12 - 1)d = 65
a + 11d = 65
(30 – 4d) + 11d = 65
[from eq (1)]
-4d + 11d = 65 - 30
7d = 35
d = 35/7
d = 5
On putting the value of d = 5 in eq (1),
a = 30 – 4d
a = 30 – 4(5)
a = 30 - 20
a = 10
By using the formula ,Sum of nth terms , Sn = n/2 [2a + (n – 1) d]
S20 = 20/2 [2(a) + (20 – 1) d]
S20 = 10 [2(10) + 19(5)]
S20 = 10 [20 + 95]
S20 = 10 × 115
S20 = 1150
Hence, the sum of first 20 terms is 1150.
HOPE THIS ANSWER WILL HELP YOU...
GIVEN :
5th term of an AP = 30
a + 4d = 30 ------(1)
12th term of an AP = 65
a + 11d = 65 ------(2)
Solve eq - 1 & 2 to find Common difference (d).
a + 4d = 30
a + 11d = 65
(-)
-------------------
-7d = -35
d = 35/7
d = 5
Common Difference = 5
Now, Substitute d in eq - (1) to find first term (a)
a + 4d = 30
a + 4(5) = 30
a + 20 = 30
a = 30 - 20
a = 10
First Term = 10
In an AP sum of the terms = n/2 ( 2a + (n - 1)d
= n/2 ( 2a + (n - 1)d
= 20/2 ( 2(10) + (20 - 1)5
= 10 ( 20 + (19)5 )
= 10 ( 20 + 95)
= 10 ( 115)
= 1150
Therefore, the sum of first 20 terms = 1150.