Question

a+b
a−b
​
+
a+b
3a−2b
​
+
a+b
5a−3b
​
+ .... up to 20 term ​

Answer 1

Step-by-step explanation:

From the given series we have

From the given series we havea ( 1st term) =a+ba−b,n=11

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)=a+b3a−2b−(a−b)=a+b3a−2b−a+b

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)=a+b3a−2b−(a−b)=a+b3a−2b−a+b⇒d=a+b2a−b

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)=a+b3a−2b−(a−b)=a+b3a−2b−a+b⇒d=a+b2a−bNow, Sn=2n[2a+(n−1)d]

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)=a+b3a−2b−(a−b)=a+b3a−2b−a+b⇒d=a+b2a−bNow, Sn=2n[2a+(n−1)d]⇒S11=211[(a+b)2(a−b)+(11−1)(a+b)(2a−b)]

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)=a+b3a−2b−(a−b)=a+b3a−2b−a+b⇒d=a+b2a−bNow, Sn=2n[2a+(n−1)d]⇒S11=211[(a+b)2(a−b)+(11−1)(a+b)(2a−b)]=2(a+b)11[2a−2b+20a−10b]

From the given series we havea ( 1st term) =a+ba−b,n=11d=(a+b)(3a−2b)−(a+b)(a−b)=a+b3a−2b−(a−b)=a+b3a−2b−a+b⇒d=a+b2a−bNow, Sn=2n[2a+(n−1)d]⇒S11=211[(a+b)2(a−b)+(11−1)(a+b)(2a−b)]=2(a+b)11[2a−2b+20a−10b]=(a+b11(11a−6b)