Question

What is the sum of all natural numbers between 100 and 300 that are divisible by 14
(A)2842
(B) 4116
(c) 1218
(D)1414
​

Answer 1

Answer:

Numbers between 100 to 300 that are divisible by 14 are

112,126,140,154,168,182,196,210,224,238,252 ,266,280, 294.

Numbers between 100 to 300 that are divisible by 14 are

112,126,140,154,168,182,196,210,224,238,252 ,266,280, 294.

This is arithmetic progression where common difference is 14, 1st term is 112, last term = 294

Total number of terms = {(last term - 1st term)/ common difference } + 1

={ (294–112)/14 } + 1

= 13 + 1 = 14 terms

Sum of AP = n(1st term + last term)/2

n is number of terms

Sum = 14(112+294)/2

14(406)/2

5684/2

2842.Ans

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Answer 2

Answer:

The correct answer is option (A) 2842

Step-by-step explanation:

To find,

The sum of all natural numbers between 100 and 300 that are divisible by 14

Solution:

Recall the formula,

The sum to n-terms of an AP = Sₙ = $\frac{n}{2}(2a+(n-1)d)$ ---------------------(1)

The number of terms in an AP = n = $\frac{a_n - a_1}{d}+1$ ---------------(2)

Here, 'a' is the 1st term of the AP, 'd' is the common difference and aₙ is the nth term of the AP

The first number between 100 and 300 divisible by 14  = 8*14 = 112

The last number between 100 and 300 divisible by 14 = 21*14 = 294

Hence the numbers between 100 and 300 divisible by 14 are

112, 126, 140, ....................294

This set form an AP, with 1st term = 112 and common difference = 14

Required to find the sum of terms of this AP

To  find the number of terms of this AP

From equation (2), we have

n = $\frac{a_n - a_1}{d}+1 = \frac{294 - 112}{14}+1$

= $\frac{182}{14} +1$

= 13+1

= 14

Hence we have,

There are 14 numbers between 100 and 300 divisible by 14

From equation (1)

Sum of 14 terms = $S_{14} = \frac{14}{2}(2a+(n-1)d)$

= 7(2× 112 +(14-1)14)

Sum of all natural numbers between 100 and 300 that are divisible by 14

=  7(224 +182)

= 7×406

= 2842

∴Sum of all natural numbers between 100 and 300 that are divisible by 14 = 2842

The correct answer is option (A) 2842

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