Question

sum of 1st 10 terms of an arithmetic sequence is 350 and sum of find 16th term is 848. write algebraic form of the sum of the sequence​

Answer 1

Answer:-

$\sf{S_{10}=350}$

$\sf{S_{16}=848}$

We know that,

$\bf{S_{n}=\dfrac{n}{2}(2a+(n-1)d)}$

Hence,

$\leadsto\sf{S_{10}=\dfrac{10}{2}(2a+(10-1)d)}$

$\leadsto\sf{350=5(2a+9d)}$

$\leadsto\sf{\dfrac{350}{5}=(2a+9d)}$

$\leadsto\sf{70=(2a+9d)}$

And,

$\leadsto\sf{S_{16}=\dfrac{16}{2}(2a+(16-1)d)}$

$\leadsto\sf{848=8(2a+15d)}$

$\leadsto\sf{\dfrac{848}{8}=(2a+15d)}$

$\leadsto\sf{106=(2a+15d)}$

Now, let's find a and d too by elimination method,

$\sf{106=(2a+15d)}$

$\sf{70=(2a+9d)}$

_________________________

$\sf{36=(6d)}$

Hence, d = 6

And, $\leadsto\sf{70=(2a+9d)}$

$\leadsto\sf{70=(2a+9(6))}$

$\leadsto\sf{70=(2a+54)}$

$\leadsto\sf{70-54=(2a)}$

$\leadsto\sf{16=(2a)}$

Hence, a = 8

$\rule{200}2$

Sum of the sequence:-

Sn = n/2 ( 2a + ( n - 1 )6 )

= n/2 ( 2( 8 ) + 6n - 6 )

= n/2 ( 16 + 6n - 6 )

= n/2 ( 10 + 6n )

= n( 5 + 3n )

= 3n² + 5n

$\rule{200}2$

Answer 2

Given :

• Sum of 1st 10 terms of an arithmetic sequence is 350.
• Sum of 1st term of the AP is 848.

To Find :

• Algebraic form of the sum of the sequence.

Solution :

Let the first term of the AP be a.

Let the common difference of the AP be d.

Case 1 :

Sum of the 1st 10 terms of the AP equals as 350.

We know that the sum of n terms of an AP is given by the formula,

$\large{\boxed{\bold{S_n\:=\:\dfrac{n}{2}\:\big[2a\:+(n-1)\:d\:\big]}}}$

Equation :

First tens terms adds up to 350.

$\longrightarrow$ $\sf{S_{10}\:=\:\dfrac{10}{2}\:\big[2a\:+\:(10-1)d\big]}$

$\longrightarrow$ $\sf{350\:=\:5\:\big[2a\:+\:(9)d\big]}$

$\longrightarrow$ $\sf{350\:=\:10a\:+\:45d}$

Divide throughout by 5,

$\longrightarrow$ $\sf{70=2a+9d\:\:\:(1)}$

Case 2 :

16 terms of the AP adds up to 848.

Equation :

$\longrightarrow$ $\sf{S_{16}\:=\:\dfrac{16}{2}\:\big[2a+\:(16-1)d\big]}$

$\longrightarrow$ $\sf{848\:=\:8\:\big[2a\:+\:(15)d\big]}$

$\longrightarrow$ $\sf{848\:=\:16a\:+\:120d}$

Divide throughout by 8,

$\longrightarrow$ $\sf{106=2a+15d\:\:(2)}$

Substract equation (1) from (2),

$\longrightarrow$ $\sf{2a+9d-(2a+15d)=70-(106)}$

$\longrightarrow$ $\sf{2a+9d-2a-15d=70-106}$

$\longrightarrow$ $\sf{9d-15d=-36}$

$\longrightarrow$ $\sf{-6d=-36}$

$\longrightarrow$ $\sf{d=\dfrac{-36}{-6}}$

$\longrightarrow$ $\sf{d=6\:\:(3)}$

Now, substitute d = 6 in equation (1),

$\longrightarrow$ $\sf{70\:=\:2a+9d}$

$\longrightarrow$ $\sf{70=2a+9(6)}$

$\longrightarrow$ $\sf{70=2a+54}$

$\longrightarrow$ $\sf{70-54=2a}$

$\longrightarrow$ $\sf{16=2a}$

$\longrightarrow$ $\sf{2a=16}$

$\longrightarrow$ $\sf{a=\dfrac{16}{2}}$

$\longrightarrow$$\sf{a=8\:\:(4)}$

Now, from (4) we have the first term of the AP, a = 8

$\longrightarrow$ $\sf{t_2\:=\:a+d}$

$\longrightarrow$ $\sf{t_2=8+6}$

$\longrightarrow$ $\sf{t_2\:=\:14}$

Third term :

$\longrightarrow$ $\sf{t_3\:=\:t_2\:+\:d}$

$\longrightarrow$ $\sf{t_3\:=\:14+6}$

$\longrightarrow$ $\sf{t_3=\:20}$

Fourth Term :

$\longrightarrow$ $\sf{t_4\:=\:t_3\:+\:d}$

$\longrightarrow$ $\sf{t_4\:=\:20+6}$

$\longrightarrow$ $\sf{t_4=\:26}$

Sequence :

$\large{\boxed{\bold{8,14,20,26,...t_n}}}$

Now, nth term of the AP will be :

$\longrightarrow$ $\sf{t_n\:=\:a\:+\:(n-1)d}$

$\longrightarrow$ $\sf{t_n\:=\:8\:+\:(n-1)6}$

$\longrightarrow$ $\sf{t_n\:=\:8+6n\:-\:6}$

$\longrightarrow$ $\sf{t_n\:=\:2\:+\:6n}$

•°• nth term of AP = 2 + 6n.

Sum of the sequence :

$\longrightarrow$ $\sf{S_n\:=\:\dfrac{n}{2}\:\big[2a\:+\:(n-1)d\big]}$

$\longrightarrow$ $\sf{S_n\:=\:\dfrac{n}{2}\:\big[2(8)\:+(n-1)\:(6)\big]}$

$\longrightarrow$ $\sf{S_n\:=\:\dfrac{n}{2}\:\big[16\:+\:6n-6\big]}$

$\longrightarrow$ $\sf{S_n\:=\:\dfrac{n}{2}\:(10\:+\:6n)}$

$\longrightarrow$ $\sf{S_n\:=\:n\:(5\:+\:3n)}$

$\longrightarrow$ $\sf{S_n\:=\:5n\:+\:3n^2}$