Special Series MCQ Quiz in मराठी - Objective Question with Answer for Special Series - मोफत PDF डाउनलोड करा

Calculations:

Consider,  =  

⇒

⇒

⇒

Hence, 

Formula Used: 

Sum of square of n numbers = 

Sum of n terms = 

Calculation:      

a₁ = 1

a₂ = 1 + 8 = 9

a₃ = 9 + 8 + 7 = 24

a₄ = 24 + 8 + 7 + 7 =  46

So, then K^th term will be:

⇒ a_k = 1 + 8(k - 1) + 7

⇒ a_k = 7k²/2 - 5k/2

Sum of first n term:

The correct option is 4 i.e.    

Concept:

Sum of first 'n' natural numbers = 

Sum of the square of first n natural numbers = 

Sum of cubes of first n natural numbers = 

Calculation:

To Find: n^th term of the series 

Concept:

• Sum of the first n Natural Numbers 
• Sum of the Square of the first n Natural Numbers 
• Sum of the Cubes of the first n Natural Numbers 

Calculation:

Here, we have to find the sum of the series 1⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + 4 ⋅ 5 +.....+ 8 ⋅ 9

As we know that, the nth term of the given series is given by T_n = n (n + 1)

Sum of series = 

As we know that,  and 

Here, n = 8

Hence, option 1 is the correct answer.

Concept:

The general term of a sequence a_n = S_{n + 1} - S_n,

Where S_n is the sum of first n terms of the series and S_{n + 1} is the sum of first (n + 1) terms of the series.

Calculation:

Given: S_n = n + 12

By substituting n by (n + 1) in the above equation, we get

S_{n + 1} = (n + 1) + 12 = n + 13

Now, General term of a sequence

⇒ a_n = S_{n + 1} - S_n = (n + 13) - (n + 12) = 1 ∀ n ∈ N

⇒ an = 1

∴ a₃ = 1

S₁ = a₁ = 1 + 12 = 13

We can write the series as 13, 1, 1, 1, 1, ............

Concept:

If a1, a2, a3,...are in GP with common ratio r, 

If a be the first term, r be the common ratio of a GP then,

Calculation:

Calculation:

Let required sum is S.

⇒ S =

⇒ S =

⇒ S = 

⇒ S = 

Threfore, a = 1/2 and r = 1/2

We know that the sum of the n^th term of GP (r

⇒ S = 

⇒ S =           (∵ 1/aⁿ = a^-n)

⇒ S = n - 1 + 2-n

∴ S = n + 2-n -1

Given:

Calculations:

Solving the given series,

After simplifying,

∴ The sum of 'n' terms of the series is 

Given:

We need to find the sum of the squares of all two-digit numbers which are completely divisible by 4.

Formula Used:

Sum of squares =

Calculation:

Two-digit numbers divisible by 4 are: 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96

Sum of squares = 12² + 16² + 20² + 24² + 28² + 32² + 36² + 40² + 44² + 48² + 52² + 56² + 60² + 64² + 68² + 72² + 76² + 80² + 84² + 88² + 92² + 96²

⇒ Sum of squares = 144 + 256 + 400 + 576 + 784 + 1024 + 1296 + 1600 + 1936 + 2304 + 2704 + 3136 + 3600 + 4096 + 4624 + 5184 + 5776 + 6400 + 7056 + 7744 + 8464 + 9216

⇒ Sum of squares = 78320

The sum of the squares of all two-digit numbers completely divisible by 4 is 78320.

Concept:

• Sum of the first n Natural Numbers 
• Sum of the Square of the first n Natural Numbers 
• Sum of the Cubes of the first n Natural Numbers 

Calculation:

Here, we have to find the sum of the series 12 ⋅ 2 + 22⋅3 + 32⋅4 + 42⋅5 +.....+up to n terms

nth term of the given series is T_n = n² ⋅ (n + 1) = n3 + n² 

Sum of n terms of the series,  

As we know that, 

 and 

Hence, option 2 is the correct answer.

Concept:

• Sum of the first n Natural Numbers 
• Sum of the Square of the first n Natural Numbers 
• Sum of the Cubes of the first n Natural Numbers 

Calculation:

Here, we have to find the sum of the series 

Let S_n = 

Multiplying the above equation by 4 we get,

⇒ 4S_n = 1 + 2 + 3 + 4 + 5 +.....+ n

As we know that, 

⇒ 4S_n = n(n + 1)/2

⇒ S_n = n(n + 1)/8

Now substitute n = 16 in the above equation we get,

⇒ S₁₆ = 17 × 2 = 34

Hence, option 4 is the correct answer.