# The 20th term of an arithmetic series is 131 and the sum of the 6th to 10th terms inclusive is 235. Find the sum of the first 20 terms. Can someone please help??

Jun 12, 2018

${S}_{20} = 1290$

#### Explanation:

$\text{in an arithmetic sequence we have}$

•color(white)(x)a_n=a+(n-1)dlarrcolor(blue)"nth term"

•color(white)(x)S_n=n/2[2a+(n-1)d]larrcolor(blue)"sum to n terms"

$\text{where a is the first term and d the common difference}$

$\text{we require to find a and d}$

$\text{here } {a}_{20} = a + 19 d = 131 \to \left(1\right)$

$\text{the sum of 6th to 10th term inclusive can be expressed as}$

$a + 5 d + a + 6 d + a + 7 d + a + 8 d + a + 9 d$

$\Rightarrow 5 a + 35 d = 235 \to \left(2\right)$

$\text{multiply "(1)" by 5}$

$5 a + 95 d = 655 \to \left(3\right)$

$\left(3\right) - \left(2\right)$

$60 d = 420 \Rightarrow d = 7$

$\text{ substitute "d=7" in } \left(1\right)$

$a + 133 = 131 \Rightarrow a = - 2$

${S}_{20} = 10 \left[\left(2 \times - 2\right) + \left(19 \times 7\right)\right] = 1290$

Jun 12, 2018

$1290$

#### Explanation:

First, let's use what we know about arithmetic sequences:

${a}_{20} = 131$

${a}_{1} + 19 d = 131 \text{ "" }$ Let's call this Equation 1.

Next, let's convert this into another equation:

${a}_{6} + {a}_{7} + \ldots + {a}_{10} = 235$

$\left({a}_{1} + 5 d\right) + \left({a}_{1} + 6 d\right) + \ldots + \left({a}_{1} + 9 d\right) = 235$

$5 {a}_{1} + \left(5 d + 6 d + 7 d + 8 d + 9 d\right) = 235$

$5 {a}_{1} + 35 d = 235$

Don't forget, we can divide by 5 to simplify.

${a}_{1} + 7 d = 47 \text{ "" }$ Let's call this Equation 2.

Now we have two equations, which we can treat as a system of equations which we then solve to find ${a}_{1}$ or $d$.

Let's subtract Equation 2 from Equation 1:

$\left({a}_{1} + 19 d = 131\right) - \left({a}_{1} + 7 d = 47\right)$

$\left(\cancel{{a}_{1}} + 19 d\right) - \left(\cancel{{a}_{1}} + 7 d\right) = 131 - 47$

$12 d = 84$

$d = 7$

Remember that ${a}_{1}$ is $19 d$ less than ${a}_{20}$, which we know is $131$. So:

${a}_{1} = {a}_{20} - 19 d = 131 - 19 \left(7\right) = - 2$

Now to find the sum of the first 20 terms, all we have to do is use the formula with the numbers we found:

${\sum}_{n = 1}^{\textcolor{b l u e}{20}} {a}_{n} = \textcolor{b l u e}{20} \cdot \left(\frac{{a}_{1} + {a}_{\textcolor{b l u e}{20}}}{2}\right)$

$= 20 \cdot \frac{\left(- 2\right) + \left(131\right)}{2}$

$= 20 \cdot \frac{129}{2}$

$= 1290$

The sum of the first 20 terms of the arithmetic sequence is $1290$.

Jun 12, 2018

${S}_{20} = 1290$

#### Explanation:

In an arithmetic series, the term is given by ${T}_{n} = a + \left(n - 1\right) d$ and the sum is given by ${S}_{n} = \frac{n}{2} \left[2 a + \left(n - 1\right) d\right]$

"The 20th term of an aritemetic series is 131
${T}_{n} = a + \left(n - 1\right) d$
${T}_{20} = 131 = a + \left(20 - 1\right) d$ ---> $131 = a + 19 d$

${S}_{10} = \frac{10}{2} \left[2 a + \left(10 - 1\right) d\right] = 5 \left[2 a + 9 d\right]$
${S}_{5} = \frac{5}{2} \left[2 a + \left(5 - 1\right) d\right] = \frac{5}{2} \left[2 a + 4 d\right]$

Now when the sum of the 6th to 10th term inclusive equals to 235 and there are 5 terms, that doesn't mean that you can do ${S}_{5}$ because that means that it is the sum of the first 5 terms, ie 1st to 5th inclusive

Instead, if it is the sum of the 6th to 10th term, you have to write instead ${S}_{10} - {S}_{5}$

${S}_{10} - {S}_{5} = 5 \left[2 a + 9 d\right] - \frac{5}{2} \left[2 a + 4 d\right] = 235$

$10 a + 45 d - 5 a - 10 d = 235$
$5 a + 35 d = 235$
$a + 7 d = 47$

$a + 19 d = 131$ --- (1)
$a + 7 d = 47$ --- (2)

(1)-(2)
$12 d = 84$
$d = 7$ --- (3)

Sub (3) into (2)
$a + 7 \left(7\right) = 47$
$a + 49 = 47$
$a = - 2$

Now, you can finally found ${S}_{20}$
${S}_{20} = \frac{20}{2} \left[2 \times - 2 + \left(20 - 1\right) \times 7\right]$
${S}_{20} = 1290$

Jun 12, 2018

Sum of first $20$ terms is $1290$

#### Explanation:

Let the $1$ st term of A.P series is $a$ and common

difference be d ; t_20= 131 ; t_20= a+(20-1)*d

 :. a +19 d = 131 (1) ; S_n=n/2 {2 a+(n-1)d}

${S}_{5} = \frac{5}{2} \left(2 a + 4 d\right) \mathmr{and} {S}_{5} = 5 a + 10 d$ similarly,

${S}_{10} = \frac{10}{2} \left(2 a + 9 d\right) \mathmr{and} {S}_{10} = 10 a + 45 d$

${S}_{10} - {S}_{5} = 235$ (given),

${S}_{10} - {S}_{5} = 10 a + 45 d - \left(5 a + 10 d\right)$ or

${S}_{10} - {S}_{5} = 5 a + 35 d = 235 \mathmr{and} a + 7 d = 47 \left(2\right)$

Subtracting equation(2) from equation(1) we get,

$12 d = 84 : d = 7 \therefore a + 19 \cdot 7 = 131 \mathmr{and} a = - 2$

:.a= -2, d=7 ; :. S_20=20/2(2*(-2)+19 *7)  or

${S}_{20} = 10 \left(- 4 + 133\right) = 10 \cdot 129 = 1290$

Sum of first $20$ terms is $1290$ [Ans]