Question

If the nth term of an AP is (2n+5),the sum of first10 terms is

Answer 1

Given :- the nth term of an AP is (2n+5) .

To Find :-

• sum of first 10 terms . ?

Solution :-

we have given that, nth term of an AP is (2n+5) .

So,

→ Tn = (2n + 5)

→ T(1) = (2 * 1 + 5) = 2 + 5 = 7 = First term.

→ T(2) = (2 * 2 + 5) = 4 + 5 = 9 = second term.

→ T(3) = (2 * 3 + 5) = 6 + 5 = 11 = Third term.

Therefore, given AP series is :- 7, 9, 11, 13 ,__________

• First term of AP = 7
• Common difference = T(2) - T(1) = 9 - 7 = 2 .

we have to find sum of first 10 terms of this AP .

we know that, sum of n terms of an AP is :-

• Sn = (n/2)[2a + (n - 1)d]

Putting values we get,

→ S(10) = (10/2)[2 * 7 + (10 - 1)2]

→ S(10) = 5[14 + 9*2]

→ S(10) = 5[14 + 18]

→ S(10) = 5 * 32

→ S(10) = 160 (Ans.)

Hence, Sum of first 10 terms of given AP is 160.

________________

Another method :-

→ Tn = (2n + 5)

→ T(1) = (2 * 1 + 5) = 2 + 5 = 7 = First term.

→ T(2) = (2 * 2 + 5) = 4 + 5 = 9 = second term.

→ T(3) = (2 * 3 + 5) = 6 + 5 = 11 = Third term.

→ T(10) = (2 * 10 + 5) = 20 + 5 = 25 = Last term.

we know that, when we have given last term of AP, we can find sum by :-

• Sn = (n/2)[First term + Last Term]

Therefore,

→ S(10) = (10/2)[7 + 25]

→ S(10) = 5 * 32

→ S(10) = 160 (Ans.)

________________________

Answer 2

SOLUTION :

GIVEN

The nth term of an AP is (2n+5)

TO DETERMINE

The sum of the first 10 terms

FORMULA TO BE IMPLEMENTED

The sum of first n natural numbers is

$\displaystyle \sf{}1 + 2 + 3 + ........ + n = \frac{n(n + 1)}{2}$

EVALUATION

Here it is given that nth term of an AP is (2n+5)

$\sf{}Let \: T_n \: be \: the \: n \: th \: term$

and

$\sf{}S_{10} = Sum \: of \: first \: 10 \: terms \: of \: the \: AP$

Then

$\sf{}T_n = 2n + 5$

Putting n = 1,2,3,...,10 we get

$\sf{}T_1 =( 2 \times 1 )+ 5$

$\sf{}T_2 =( 2 \times 2 )+ 5$

$\sf{}T_3 =( 2 \times 3 )+ 5$

.

.

.

$\sf{}T_{10} =( 2 \times 10 )+ 5$

-----------------------------------------------

Adding we get

$\sf{}T_1+T_2+T_3+......+T_{10}$

$= \sf{}2(1 + 2 + 3 + .... + 10) +( 5 \times 10 )$

$\displaystyle \implies \sf{}S_{10} = 2 \times \frac{10(10 + 1)}{2} + 50$

$\displaystyle \implies \sf{}S_{10} = {10(10 + 1)} + 50$

$\displaystyle \implies \sf{}S_{10} = 110 + 50$

$\displaystyle \implies \sf{}S_{10} = 160$

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