The 19th and 22nd term of an AP are 29 and 55 respectively. find the sum of the 60th term
a + 18d = 29
a + 21d =55
subtract:
3d = 26
d = 26/3
a + 18(26/3) = 29
a = -127
"find the sum of the 60th term" makes no sense
Do you want the 60th term of the sum of the first 60 terms?
either way, just plug in the values of a and d into the formulas
the 19th and 22nd terms differ by 3d.
So, d=(55-29)/3 = 8
Now you can find a
Not sure what the sum of the 60th term means, but the sum of the first 60 terms is
s = 60/2 (2a+59d)
go with Reiny. I assumed an integer value for d, and missed the fact that 55-29=26, not 24.
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26
To find the sum of the 60th term in an arithmetic progression (AP), we need to first find the common difference (d) of the AP.
We can calculate the common difference (d) as follows:
d = 22nd term - 19th term
= 55 - 29
= 26
Now, we have the common difference (d) as 26. Let's find the first term (a) of the AP.
To find the first term (a), we can use the formula:
a = 19th term - (n-19)d
Substituting the known values into the formula:
29 = a - 26(19 - 19)
29 = a
Now, we have the first term (a) as 29 and the common difference (d) as 26.
To find the sum of the 60th term (Sn), we can use the formula:
Sn = (n/2)(2a + (n-1)d)
Substituting the known values into the formula:
S60 = (60/2)(2 * 29 + (60-1) * 26)
= 30(58 + 59 * 26)
= 30(58 + 1534)
= 30(1592)
= 47,760
Therefore, the sum of the 60th term is 47,760.