Question : What is the sum of all natural numbers between 1 and 100 which are multiples of 7?
Option 1: $735$
Option 2: $675$
Option 3: $745$
Option 4: $705$
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Correct Answer: $735$
Solution :
The numbers between 1 and 100 which are divisible by 7 are (7,14,21..........98)
Equation for the number of terms: $T_n = a + (n–1)d$
Where $a$ = 1
st
term
$n$ = number of terms
$d$ = difference
Putting the values we get,
$98 = 7 + (n - 1)7$
⇒ $91 = (n-1)7$
⇒ $n = 14$
Now sum of terms $S = \frac{n}{2}(2a+(n-1)d)$
= $\frac{14}{2}(2\times7+(14–1)7)$
= $\frac{14}{2}(14+(13×7))$
= $7 × (14 + 91)$
= $7 × 105$
= $735$
Hence, the correct answer is $735$.
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