An arithmetic sequence has its fifth term equal to 22 and its 15 term is equal to 62. find its hundredth term and sum of first 50 terms
Answers
Given that,
a5 = 22
a + 4d = 22 ---(1)
Same here, a15 = 62
a + 14d = 62 ---(2)
using equations (1) and (2), we obtain
10d = 40
d = 4
Add the value of d in equation .(1),
a + 4(4) = 22
a = 22 - 16 = 6
Thus, a = 6
Now, a = 6
n = 100
d = 4
We know that, an
= a + (n -1)d
a100 = 6 +(100 - 1)× 4
a100 = 6 + 99×4
= 6 + 396 = 402
it's 100th term = 402
Sum of first 50 terms
= n/2[2a +(n -1)d]
=50/2[ 2×6 + (50 -1)× 4]
= 25[ 12 + 49× 4]
= 25[ 12+196]
= 25× 208 = 5200Ans
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Given:
Fifth term of the AP = 22
Fifteenth term of the AP = 62
5 th term of the AP can be written as:
a + 4d = 22---------Equation 1
15 th term of the Ap can be written as:
a + 14d = 62-----------Equation 2
Solving both the equation by Elimination method:
a + 4d = 22
a + 14d = 62
- 10 d = - 40
d = - 40 / - 10
d = 4
Substituting the value of d in Equation 2:
a + 14 x 4 = 62
a + 56 = 62
a = 62 - 56
a = 6
Calculating the 100th term of the AP:
= a + 99d
Substituting the values we know into this equation:
= 6 + 99 x 4
= 6 + 396
= 402
Therefore, the 100 th term of the AP is 402.
Now calculating the sum of the first 50 terms of the ap:
Sn = n/2(2a + (n - 1) d)
Substituting the values we know into this formula:
= 50 / 2 (2 x 6 + (50 - 1) 4
= 25 x (12 + 196)
= 25 x 208
= 5200
Therefore, the sum of the first 50 terms is 5200.