Question

the n term of an a.p is 3n²-n then sum of the first 10 term will be:
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Answer 1

Step-by-step explanation:

Given :-

The nth term of an AP is 3n²-n.

To find :-

Find the sum of the first 10 terms ?

Solution :-

Given that

The nth term of an AP is 3n²-n.

an = 3n²-n ---------(1)

Put n = 1 in (1) then

a1 = 3(1)²-1

=> a1 = 3(1)-1

=> a1 = 3-1

=> a1 = 2

First term = 2

Put n = 2 in (1) then

=> a2 = 3(2)²-2

=> a2 = 3(4)-2

=> a2 = 12-2

=> a2 = 10

Second term = 10

Common difference =a2-a1

=> d = 10-2

=> d = 8

We have , a = 2 , d = 8

We know that

The sum of the first n terms in an AP

Sn = (n/2)[2a+(n-1)d]

Sum of the first 10 terms

=> S10 = (10/2)[2(2)+(10-1)(8)]

=> S10 = (5)[4+9(8)]

=> S10 = (5)(4+72)

=> S10 = 5(76)

=> S10 = 380

Answer:-

The sum of the first 10 terms of the given AP is 380

Used formulae:-

→ The sum of the first n terms in an AP

Sn = (n/2)[2a+(n-1)d]

• a = First term
• d = Common difference
• n = Number of terms

Answer 2

SOLUTION

GIVEN

The nth term of an AP is 3n² - n

TO DETERMINE

The sum of the first 10 term

FORMULA TO BE IMPLEMENTED

We are aware of the formula that

$\displaystyle \sf\sum\limits_{r=1}^{n} r= \frac{n(n + 1)}{2}$

$\displaystyle \sf\sum\limits_{r=1}^{n} {r}^{2} = \frac{n(n + 1)(2n + 1)}{6}$

EVALUATION

Here it is given that nth term of an AP is 3n² - n

So by the given condition

$\sf \: a_n = 3 {n}^{2} - n$

Hence the required sum of the first 10 term

$\sf = S_{10}$

$\displaystyle = \sf\sum\limits_{n=1}^{10} a_n$

$\displaystyle = \sf\sum\limits_{n=1}^{10} (3 {n}^{2} - n )$

$\displaystyle = \sf3\sum\limits_{n=1}^{10} {n}^{2} - \sf\sum\limits_{n=1}^{10} n$

$\displaystyle \sf= 3 \times \frac{10(10 + 1)(20 + 1)}{6} - \frac{10(10 + 1)}{2}$

$\displaystyle \sf= \frac{10 \times 11 \times 21}{2} - \frac{10 \times 11}{2}$

$\displaystyle \sf= 1155 - 55$

$\displaystyle \sf= 1100$

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