Question

In AP sum of the first 10 term is 155 & sum of the first 20 term is 610 then find the sum of the term from 11th term to 20th term

Answer 1

GiveN :

• Sum of 10 terms is 155
• Sum of 20 terms is 610

To FinD :

• Sum of the terms from 11th term to 20th term

Solution :

In this question we are given the sum of first 10 terms which is 155 ,and also the sum of first 20 terms which is 610. and we have to calculate the value of sum of the 11th term to the 20th term .

In this case we can directly subtract the value of sum of first 10 terms from the sum of first 20 terms. There is no need of doing calculations.

$\rule{200}{2}$

A.T.Q,

let the sum of term from 11th to 20th be Sn.

$\longrightarrow \sf{S_n \: = \: S_{20} \: - \: S_{10}} \\ \\ \longrightarrow \sf{S_n \: = \: 610 \: - \: 155} \\ \\ \longrightarrow \sf{S_n \: = \: 455} \\ \\ \underline{\underline{\sf{Sum \: of \: the \: terms \: from \: 11th \: to \: 20th \: term \: is \: 455}}}$

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Some Additional Information :

• In an AP, the difference between the two consecutive terms should be equal.

• The first term of AP is denoted by a

• Add the next terms by adding number of terms in subscript. Example second term is denoted by $\sf{a_2}$

• And formula for calculating the last term is $\sf{a_n \: = \: a \: + \: (n \: - \: 1)d}$

Where,

• an is last term

• a is first term

• n is number of terms

• d is difference

Answer 2

Answer:

455

Step-by-step explanation:

Given:

Sum of first 10 terms of the given Arithmetic progression = 155

Sum of first 20 terms of the Arithmetic progression = 610

Sum of n terms of an A.P is given by the formula:

$\frac{n}{2} (2a + (n - 1)d)$

Where n = number of terms

a = first term

d = common difference

Substituting the values, we get:

$\frac{10}{2} (2 \times a + (10 - 1)d)$

$10a + 9d$

Similarly for the sum of first 20 terms, we get:

$\frac{20}{2} (2 \times a + (20 - 1)d)$

$20a + 19d$

We know that 10a + 9d = 155, 20a + 19d =610

To find the sum of terms between 20th and 11th term, we need to subtract the (sum of 20 terms of the A.P - sum of the 10 terms of the A.P)

Sum of terms from 11th term to 20th term = 610-155

= 455

In relative to the above expressions we can say :

455= 20a+19d-(10a+9d)

455= 10a + 10d

The sum is 455