Question

Find the sum of the first 50 common terms of 12,16,20..... and 18,24,30...

Answer 1

Answers will be
5500 & 8250

Answer 2

Answer:

Given the series : 12, 16, 20 ,.......

Since the given series is an Arithmetic series as common difference between the numbers is constant i.e, 4

To find the sum of the first 50 terms of this series is;

First find the 50th terms

Using formula:

$a_n =a+(n-1)d$ where a is the first term , n is the number of terms and d is the common difference

Here, n = 50 ,common difference(d) = 16-12=20-16 = 4 and a=12

then;

$a_{50}=12+(50-1)(4)$ or

$a_{50}=12+49 \cdot 4 = 208$

Now, to find the sum of first n terms of an arithmetic sequence we use the formula:

$S_n= \frac{n (a+a_n)}{2}$ ; where n is the number of terms, a is the first term and $a_n$ be the last term of the series.

Using this formula to get the sum of 50 terms:

$S_{50}=\frac{50(12+208)}{2} = \frac{50 \cdot 220}{2} =50 \cdot 110 = 5500$.

Similarly, find the sum of first 50 terms of 18,24,30, ......

Since the given series is an Arithmetic series as common difference between the numbers is constant i.e, 6

To find the sum of the first 50 terms of this series is;

First find the 50th terms;

Using formula:

$a_n =a+(n-1)d$ where a is the first term , n is the number of terms and d is the common difference

Here, n = 50 ,common difference(d) = 24-18=30-24 = 6 and a=18

then;

$a_{50}=18+(50-1)(6)$ or

$a_{50}=18+49 \cdot 6 = 18+ 294=312$

Now, to find the  sum of first n terms of an arithmetic sequence we use the formula:

$S_n= \frac{n (a+a_n)}{2}$ ; where n is the number of terms, a is the first term and $a_n$ be the last term of the series.

Using this formula to get the sum of 50 terms:

$S_{50}=\frac{50(18+312)}{2} = \frac{50 \cdot 330}{2} =50 \cdot 165 = 8250$.

Therefore, the Sum of the first 50 common terms of 12,16,20..... and 18,24,30... is = 5500+8250 = 13,750