Question

consider 5,10,15....what is the difference between first 20 terms and the next 20 terms​

Answer 1

2,000

Step-by-step explanation:

The given series is 5, 10, 15.... Let us check whether this series is an arithmetic progression (A.P) or not by finding the common difference.

We know that,

⟹ Common difference = Second Term - First term

⟹ Common difference = 10 - 5

⟹ Common difference = 5

Also,

⟹ Common difference = Third Term - Second Term

⟹ Common difference = 15 - 10

⟹ Common difference = 5

Therefore, the given series is in A.P in which a = 5, d = 5.

Finding sum of first 20 terms:

$\implies{S_n = \dfrac{n}{2}[2a+(n-1)d]}$

$\implies{S_{20} = \dfrac{20}{2}[2 \times 5+(20-1) \times 5]}$

$\implies{S_{20} = 10[10+19 \times 5]}$

$\implies{S_{20} = 10[10+95]}$

$\implies{S_{20} = 10 \times 105}$

$\implies{S_{20} = 1050}$

Next, we have to find the sum of next 20 terms. This implies that, we need to find the sum from 21st term to 40th term.

$\implies{a_{21} = a+20d}$

$\implies{a_{21} = 5+20 \times 5}$

$\implies{a_{21} = 5+100}$

$\implies{a_{21} = 105}$

Similarly,

$\implies{a_{40} = a+39d}$

$\implies{a_{40} = 5+39 \times 5}$

$\implies{a_{40} = 5+195}$

$\implies{a_{40} = 200}$

Since we are finding the sum from 21st to 40th term, the first term(a) = 20th term (105). The last term(l) = 40th term (200)

Therefore, sum of next 20 terms is:

$\implies{S_n = \dfrac{n}{2}[a+l]}$

Here, n = 20, a = 105, l (Last term) = 200.

$\implies{S_{20} = \dfrac{20}{2}[105+200]}$

$\implies{S_{20} = 10[105+200]}$

$\implies{S_{20} = 10 \times 305}$

$\implies{S_{20} = 3050}$

Hence, According to the question:

Sum of next 20 terms - Sum of first 20 terms

= 3050 - 1050

= 2000

Answer 2

Given :- consider 5,10,15....what is the difference between sum of first 20 terms and the next 20 terms ?

Solution :-

given AP series is :- 5, 10, 15, ______

so,

→ first term = a = 5

→ common difference = 10 - 5 = 5

then,

→ Sn = (n/2)[2a + (n - 1)d]

→ S(20) = (20/2)[2*5 + (20-1)5]

→ S(20) = 10[10 + 95]

→ S(20) = 5 * 105 = 1050

and,

→ S(40) = (n/2)[2a + (n - 1)d]

→ S(40) = (40/2)[2*5 + (40-1)5]

→ S(40) = 20[10 + 195]

→ S(40) = 20 * 205 = 4100

then,

→ Sum of next 20 terms = 4100 - 1050 = 3050 .

therefore,

→ Required difference = 3050 - 1050 = 2000 (Ans.)

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