An important question from ML Aggarwal It is given that the third term of an ap is 1 and 6th term is -11, find the sum of its first 32term.
Arithmetic Progression Chapter 9 Question no 10
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If the third term of an ap is 1 and 6th term is -11find the sum of its first 32term
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Given,
a3=1
a6=−11
nth term of A.P. is calculated as,
an=a1+(n−1)d
For 3rd term, put n=3
∴a3=a1+(3−1)d
∴1=a1+(2)d
∴a1+2d=1 Equation (1)
For 6th term, put n=6
∴a6=a1+(6−1)d
∴−11=a1+(5)d
∴a1+5d=−11 Equation (2)
Subtract equation (1) from (2), we get,
(a1−a1)+(5d−2d)=−11−1
3d=−12
∴d=−4
Put this value in equation (1), we get,
a1+2(−4)=1
∴a1−8=1
∴a1=9
Thus, sum of first 32 terms is calculated by,
∑an=2n[2a+(n−1)d]
∴∑an=232[(2×9)+(32−1)(−4)]
∴∑an=16[18+31(−4)]
∴∑an=16[18−124]
∴∑an=16[−106]
∴∑an=−1696