Question

If the sum of first n terms of an AP is given by n² + 6n, then the 10th term of the AP will be?​

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Sum of first "n" terms = n² + 6n.

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• We need to find the "10th" term.

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

We already know that sum:

$\boxed{\sf{S_n=n^2+6n}}$

Where n is the number of terms.

Do not just simply substitute the number of terms (n) = 10.

Firstly, to find the first term, that is, (a), you substitute n=1.

$\sf S_1=1^2+6$

$\sf S_1=7$

And, now S₂ would be:

$\sf S_2=2^2+6 \times 2$

$\sf S_2=16$

And, remember that S₁=a. Therefore, a=7.

And S₂ = a + a₂. [S is the sum of the terms.]

⇒ 16 = 7 + a₂

⇒ a₂ = 11.

Common Difference = a₂ - a = 11 - 7 = 4.

Therefore, we found out that:

• a = 7
• d = 4
• And it is given in the question that the 10th term, i.e., n=10.

Now use the formula.

$\to \sf a_n=a+(n-1)d$

$\to \sf a_{10}=7+(10-1)4$

$\to \sf a_{10}=7+36$

$\leadsto \sf{\red{a_{10}=43}}$

-Therefore, the 10th term is 43.

Answer 2

Given :-

• Sum of first "n" terms = n² + 6n.

What to do :-

• We have to find the "10th" term.

Solution :-

We already know that sum :-

$\boxed{ \: \: \: \: \: \huge \bold \pink{S_{n} = {n}^{2} + 6n \: \: \: \: }}$

• Where n is the number of terms.

• Do not just simply substitute the number of terms (n) = 10

Firstly,

• To find the first term, that is, (x), you substitute n=1.

$\bold{S_{1} = 1 {}^{2} + 6 }$

$\bold{S_{1} = 7}$

------------And----------

• Now S₂ would be :-

$\bold{S_{2} = 2 {}^{2} + 6 \times 2}$

$\bold{S_{2} = 16}$

------------And---------------

• Remember that S₁=a.

Therefore,

• a = 7

• And S₂ = a + a₂. [S is the sum of the terms.]

$\bold{⇒ 16 = 7 + a₂}$

$\bold{⇒ a₂ = 11.}$

• Common Difference = a₂ - a = 11 - 7 = 4.

Therefore,

We found out that :-

• a = 7

• Difference (d) = 4

----------And--------

• It is given in the question that the 10th term, i.e., n=10.

Now use the formula :-

$\bold \red{ a_{n} = a + (n - 1)d}$

$\bold \red{ a_{10} = 7 + (10 - 1)4}$

$\bold \red{ a_{10} = 7 + 36}$

$\bold \orange{ a_{10} = 43 }$

→ Hence,

• The 10th term is 43.

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