Question

If sum to n terms of an A.P. is 5n2 +6n and rth term
is 401, then the value of r is​

Answer 1

Answer:

Given:

• $\sf{S_n = 5n^2 + 6n}$
• $\sf{a_r = 401}$

$\mathfrak{\underline{Solution:-}}$

Using the relation for sum to n terms in an AP , we can find first term (a) and common difference (d) and hence rth term

→$\sf{S_n = 5n^2 + 6n}$

Putting n = 1 ,

→$\sf{S_1 = 5(1)^2 + 6(1)}$

→$\sf{a = 5 + 6}$

→$\sf{a = 11}$

Similarly putiing n = 2 ,

→$\sf{S_2 = 5(2)^2 + 6(2)}$

→$\sf{S_2 = 20 + 12}$

→$\sf{S_2 = 32}$

Now , we can find the second term , using formula :

$\boxed{\sf{a_n = S_n - S_{n-1}}}$

Putting n = 2 here, :

➫ $\sf{a_2 = S_2 - S_{2-1}}$

➫ $\sf{a_2 = 32 - S_1}$

➫ $\sf{_2 = 32 - 11}$

➫ $\sf{a_2 = 21}$

Now we can find d using the formula:

$\boxed{\sf{d = a_2 - a_1}}$

➫$\sf{d = a_2 - a_1}$

➫$\sf{d = 21 - 11}$

➫$\sf{d = 10}$

Now , r can be found using the relation:

$\boxed{\sf{a_r = a + (r-1)d}}$

➫$\sf{401 = 11 + (r-1)10}$

➫$\sf{401 - 11 = (r-1)10}$

➫$\sf{\dfrac{390}{10} = r-1}$

➫$\sf{r- 1 = 39}$

➫$\sf{r = 40}$

•°•Value of r is 40

Answer 2

Answer:

• Value of r is 40

Explanation:

Given,

• Sum of n terms of an AP is 5 n² + 6 n
• rth term of AP =  aᵣ = 401

To find,

• Value of r

So, Finding the AP

→ First term of AP = a₁ = S₁ = 5 ( 1 )² + 6 ( 1 ) = 5 + 6 = 11

→ Second term of AP = a₂ = S₂ - a₁ = [ 5 ( 2 )² + 6 ( 2 ) ] - 11 = 20 + 12 - 11 = 21

→ Third term of AP = a₃ = S₃ - ( a₁ + a₂ ) = [ 5 ( 3 )² + 6 ( 3 ) ] - ( 11 + 21 ) = 45 + 18 - 32 = 31

Hence, Our AP is : 11 , 21 , 31 , ..........

Now,

• First term of AP, a = 11

• common difference of AP, d = 21 - 11 = 10

As given that rth term of AP is 401

so,

→ aᵣ = a + ( r - 1 ) d

→ 401 = ( 11 ) + ( r - 1 ) ( 10 )

→ 401 = 11 + 10 r - 10

→ 401 = 1 + 10 r

→ 10 r = 400

→ r = 40

therefore,

• Value of r is 40.