if the sum of first 7 term of an AP is 49 and that of 17 terms os 289, find the sum of the nth terms
Answers
Answer:
n*n
Step-by-step explanation:
Here consider,
Sum of 7 terms of an A.P=49
By using formula which is,
S= n/2*[2a+(n-1)d]
49= 7/2*[2a+(7-1)d]
49=7/2*[2a+6d]
49=7*[a+3d] (we get a+3d by dividing 2a+6d by 2)
49= 7a+21d (divide the terms by 7 on both sides)
7= a+3d --1
Step 2: form the equation using S= n/2[2a+(n-1)d]
Sum of 17 terms = 289
289=17/2*[2a+(17-1)d]
289= 17/2 *[2a+16d]
289= 17*[a+8d] (we get a+8d by dividing 2a+16d by 2)
289 = 17a+136d (divide the terms by 17 on both sides)
17 = a+8d --2
From 2 and 1
17 = a+8d
(-) 7= a+3d (here when subtracted on both sides we need to take - as
10 = 5d common)
d=10/5, therefore d = 2
we got d=2, we can keep the d value using any equation, so
a+3d=7 a+8d =17
=> a+3*(2)=7 or => a+8*(2)=17
=> a+6=7 => a+16=17
=> a=7-6 => a=17-16
=> a= 1 = > a= 1
now we got a=1 and d=2, so
we need to use the formula which is
S=n/2[2a+(n-1)d]
use the values a=1 and d=2
S=n/2[2a+(n-1)*d]
S= n/2[2*(1)+(n-1)*2]
S= n/2[2+2n-2]
S= n/2[2n] (since 2-2=0, therefore there will be only 2n inside [])
S= n*[n] (here 2n is divided by 2)
Therefore S= n*n or n2