the sum of the first five terms of an arithmetic sequence is 105 and the sum of first ten terms is 385 what is the third term of the sequence
Answers
Question :
The sum of the first five terms of an arithmetic sequence is 105 and the sum of first ten terms is 385. What is the third term of the sequence?
Answer :
Third term of AP = 21
Formula used :
nth term of AP, a_n = a + (n - 1)d
where
a = first term
n = no. of term
d = common difference
Solution :
Given,
In an AP,
Sum of first five terms = 105
→ a_1 + a_2 + a_3 + a_4 + a_5 = 105
→ a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 105
→ 5a + 10d = 105
→ a + 2d = 21 -------- equation 1
Sum of first 10 terms = 385
→ a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8 + a_9 + a_10 = 385
→ 105 + (a + 5d) + (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) = 385
→ 5a + 35d = 280
→ a + 7d = 56 ------- equation 2
Subtract equation 1 from 2
a + 7d = 56
a + 2d = 21 (-)
_____________
5d = 35
_____________
→ d = 7
From equation 1,
a + 2d = 21
a + 2(7) = 21
→ a = 7
Third term of the AP,
a_3 = a + 2d
a_3 = 7 + 2(7)
a_3 = 21
Therefore,
Third term of the AP = 21
Answer:
21
Step-by-step explanation:
let a,a+d,a+2d,a+3d,a+4d be 1st 5 terms,
given ,
a+ a+d + a+2d + a+3d + a+4d = 105,
=> 5a + 10d = 105,------>eqn1,
similarly,
sum of 1st 10 terms means,
a+ a+d + a+2d + ........ + a+9d = 385,
=> 10a + 45d = 385,------->eqn2,
solving eqn1 & eqn 2,
we get a= 7 , d=7.
Then 3rd term will be,
tn = a + (n-1)d,
T3 = 7+(3-1)7 = 21.