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??

4 Answers
Jun 12, 2018

S_20=1290

Explanation:

"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"

"where a is the first term and d the common difference"

"we require to find a and d"

"here "a_20=a+19d=131to(1)

"the sum of 6th to 10th term inclusive can be expressed as"

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

rArr5a+35d=235to(2)

"multiply "(1)" by 5"

5a+95d=655to(3)

(3)-(2)

60d=420rArrd=7

" substitute "d=7" in "(1)

a+133=131rArra=-2

S_20=10[(2xx-2)+(19xx7)]=1290

Jun 12, 2018

1290

Explanation:

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

a_20 = 131

a_1 + 19d = 131 " "" " Let's call this Equation 1.

Next, let's convert this into another equation:

a_6 + a_7 + ... + a_10 = 235

(a_1 + 5d) + (a_1 + 6d) + ... + (a_1 + 9d) = 235

5a_1 + (5d+6d+7d+8d+9d) = 235

5a_1 + 35d = 235

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

a_1 + 7d = 47 " "" " 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:

(a_1 + 19d = 131) - (a_1 + 7d = 47)

(cancel(a_1) + 19d) - (cancel(a_1) + 7d) = 131 - 47

12d = 84

d = 7

Remember that a_1 is 19d less than a_20, which we know is 131. So:

a_1 = a_20 - 19d = 131 - 19(7) = -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)^color(blue)20 a_n = color(blue)20 * ((a_1 + a_color(blue)20)/2)

= 20 * ((-2) + (131) )/ 2

= 20 * 129/2

= 1290

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

Final Answer

Jun 12, 2018

S_20=1290

Explanation:

In an arithmetic series, the term is given by T_n=a+(n-1)d and the sum is given by S_n=n/2[2a+(n-1)d]

"The 20th term of an aritemetic series is 131
T_n=a+(n-1)d
T_20=131=a+(20-1)d ---> 131=a+19d

S_10=10/2[2a+(10-1)d]=5[2a+9d]
S_5=5/2[2a+(5-1)d]=5/2[2a+4d]

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[2a+9d]-5/2[2a+4d]=235

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

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

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

Sub (3) into (2)
a+7(7)=47
a+49=47
a=-2

Now, you can finally found S_20
S_20=20/2[2times-2+(20-1)times7]
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=5/2(2 a+4 d) or S_5= 5 a +10 d similarly,

S_10=10/2(2 a+9 d) or S_10= 10 a +45 d

S_10-S_5 = 235 (given),

S_10-S_5 =10 a +45 d-( 5 a +10 d ) or

S_10-S_5 =5 a + 35 d= 235 or a +7 d = 47 (2)

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

12 d= 84 : d= 7 :. a+ 19*7 = 131 or a = -2

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

S_20=10(-4+133) = 10 *129=1290

Sum of first 20 terms is 1290 [Ans]