the 9th and 22nd terms of an A.P are 29 and 55,respectively.find the sum of its first 10 term
yeah
yes
13d = 55-29
d = 2
so a = 29-8*2 = 13
S10 = 10/2 (2a+9d)
To find the sum of the first 10 terms of an arithmetic progression (A.P.), we first need to find the common difference, d.
Given that:
The 9th term, a₉, is 29, and
The 22nd term, a₂₂, is 55.
We can use the formula for the nth term of an A.P. to find the common difference:
aₙ = a + (n - 1)d
Substituting the values given, we have:
a₉ = a + 8d (1)
a₂₂ = a + 21d (2)
Subtracting equation (1) from equation (2), we eliminate 'a' and find:
a₂₂ - a₉ = 21d - 8d
55 - 29 = 13d
26 = 13d
d = 2
Now that we know the common difference, we can find the sum of the first 10 terms of the A.P using the formula:
Sum of first n terms = (n/2)(2a + (n-1)d)
Substituting the values into the formula:
n = 10 (since we want to find the sum of the first 10 terms)
a = the first term of the A.P (a₁)
d = common difference
We do not know a₁, but we can find it using the formula for the nth term of an A.P.:
aₙ = a + (n-1)d
Substituting the known values:
29 = a + 8(2)
29 = a + 16
a = 29 - 16
a = 13
Now we can find the sum of the first 10 terms:
Sum = (n/2)(2a + (n-1)d)
= (10/2)(2(13) + (10-1)(2))
= (5)(26 + 9(2))
= (5)(26 + 18)
= (5)(44)
= 220
Therefore, the sum of the first 10 terms of the given A.P. is 220.