0 [email protected]Guru Asked: March 16, 20212021-03-16T09:19:39+05:30 2021-03-16T09:19:39+05:30In: ICSE If the third term of an ap is 1 and 6th term is -11find the sum of its first 32term 0 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 arithmetic progressionicseml aggarwal Share Facebook 1 Answer Voted Oldest Recent MathsMentor Guru 2021-03-16T19:00:41+05:30Added an answer on March 16, 2021 at 7:00 pm 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 0 Reply Share Share Share on Facebook Share on Twitter Share on LinkedIn Share on WhatsApp Leave an answerLeave an answerCancel reply Featured image Select file Browse Add a Video to describe the problem better. Video type Youtube Vimeo Dailymotion Facebook Choose from here the video type. Video ID Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs". Click on image to update the captcha. Save my name, email, and website in this browser for the next time I comment. Related Questions Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + ... Problem 2: Find the indicated terms in each of the following sequences whose nth terms are(iii) an = ... Problem 1: Write the first terms of each of the following sequences whose nth term is: (i) an ... Ques (b) In the figure (ii) given below, O and O’ are centres of two circles touching each ... Question 39. (a) In the figure (i) given below, AB is a chord of the circle with centre ...

## 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