0 Rajan@2021Guru Asked: March 16, 20212021-03-16T15:48:07+05:30 2021-03-16T15:48:07+05:30In: ICSE Find the sum given below. (i) 34+32+30+….+10. (ii) (−5)+(−8)+(−11)+…..+(−230) 0 We have been asked to calculate the sum given below. (i) 34+32+30+….+10. (ii) (−5)+(−8)+(−11)+…..+(−230) ML Aggarwal Avichal Publication Arithmetic Progression chapter 9 question no 2 arithmetic progressionicseml aggarwal Share Facebook 1 Answer Voted Oldest Recent MathsMentor Guru 2021-03-16T15:56:34+05:30Added an answer on March 16, 2021 at 3:56 pm (i) 34,32,30....................10 Here a=34,l=10 d=32−34=−2 an=a+(n−1)d 10=34+(n−1)(−2)⇒10−34=(−2)(n−1)⇒n−1=12⇒n=13 and last term l=10 We know that Sn=2n(a+l) Sn=213(34+10) =213×44 =286 (ii) −5+(−8)+(−11)+...........+(−230) Here a=−5,l=−230 d=(−8−(−5))=−3 an=a+(n−1)d 230=−5+(n−1)(−3)⇒−230+5=(−3)(n−1)⇒n−1=75⇒n=76 and last term l=−230 We know that Sn=2n(a+l) Sn=276(−5+(−230) =−8930 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 ...

(i) 34,32,30....................10

Here a=34,l=10

d=32−34=−2

an=a+(n−1)d

10=34+(n−1)(−2)⇒10−34=(−2)(n−1)⇒n−1=12⇒n=13

and last term l=10

We know that Sn=2n(a+l)

Sn=213(34+10)

=213×44

=286

(ii) −5+(−8)+(−11)+...........+(−230)

Here a=−5,l=−230

d=(−8−(−5))=−3

an=a+(n−1)d

230=−5+(n−1)(−3)⇒−230+5=(−3)(n−1)⇒n−1=75⇒n=76

and last term l=−230

We know that Sn=2n(a+l)

Sn=276(−5+(−230)

=−8930