Question

The algebra of sum of an arithmetic sequence is [n]2+2n. a.) What is the sum of first 10 term of this sequence. b.) How many terms from the first ate needed to get the sum 168?

Answer 1

Rule for the sum of arithmetic sequence=n²+2 n

As we know sum of n terms of an Arithmetic sequence is $\frac{n}{2}[2 a+(n-1)d]$

⇒n²+2 n=$\frac{n}{2}[2 a+(n-1)d]$

⇒2(n+2)=2 a+ (n-1) d

⇒2 n + 4=2 a-d + n d

Equating LHS and RHS

⇒2 n = n d  and 2 a -d= 4

⇒ d=2  ∧  2 a- 2=4

⇒2 a= 6

⇒a=3

a). Sum of first 10 term of this sequence, put n=10 in $S_{n}$

$S_{n}$ =  n²+ 2 n=10²+2×10=100+20=120

(b) Let p terms are needed to get the sum 168.

⇒p²+ 2 p=168

⇒p²+ 2 p-168=0

⇒(p+14)(p-12)=0

⇒p≠ -14[ number of terms can't be negative]

So, p=12

Answer 2

Rule for the sum of arithmetic sequence=n²+2 n

As we know sum of n terms of an Arithmetic sequence is  n/2(2a + (n-1)d)

⇒n²+2 n=  n/2(2a + (n-1)d)

⇒2(n+2)=2 a+ (n-1) d

⇒2 n + 4=2 a-d + n d

Equating LHS and RHS

⇒2 n = n d  and 2 a -d= 4

⇒ d=2  ∧  2 a- 2=4

⇒2 a= 6

⇒a=3

a). Sum of first 10 term of this sequence, put n=10 in  

Sn =  n²+ 2 n=10²+2×10=100+20=120

(b) Let p terms are needed to get the sum 168.

⇒p²+ 2 p=168

⇒p²+ 2 p-168=0

⇒(p+14)(p-12)=0

⇒p≠ -14[ number of terms can't be negative]

So, p=12