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Chapter 11: Q6. (page 586)

Find Sn for each arithmetic series described.

localid="1648548029213" a1=132,d=鈭�4,an=52

Short Answer

Expert verified

Sn for the arithmetic series isS21=1932 .

Step by step solution

01

Step 1. Given Information.

Given arithmetic series is a1=132,d=鈭�4,an=52.

02

Step 2. Calculation.

The nth term of an arithmetic series is given by an=a1+(n鈭�1)d

Here, a1=132,d=鈭�4,an=52a1=132,d=鈭�4,an=52

Plugging the values:

an=a1+(n鈭�1)d52=132+(n鈭�1)(鈭�4)52鈭�132=(n鈭�1)(鈭�4)(n鈭�1)(鈭�4)=鈭�80n鈭�1=20n=20+1n=21

The sum Sn of the first n terms of an arithmetic series is given by

Sn=n2(2a1+(n鈭�1)d)

Here, a1=132,n=21,d=鈭�4a1=132,n=21,d=鈭�4

Plugging the values:

Sn=n2(2a1+(n鈭�1)d)S21=212(2(132)+(21鈭�1)(鈭�4))S21=212(264+(20)(鈭�4))S21=212(264鈭�80)S21=212(184)S21=21(92)S21=1932

03

Step 3. Conclusion.

Hence, the sum isS21=1932.

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