In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP,  Wh

Subject

Mathematics

Class

CBSE Class 10

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8.

In an AP, if S5 + S7 = 167 and S10 = 235, then find the AP,  Where Sn denotes the sum of its first n terms.


Let a and d be the first term and the common difference of the AP, respectively

therefore space Sum space of space straight n space terms comma space straight S subscript straight n space equals space straight n over 2 space left square bracket 2 straight a space plus space left parenthesis straight n minus 1 right parenthesis straight d right square bracket
We space have
straight S subscript 5 space plus space straight S subscript 7 space equals space 167

rightwards double arrow space 5 over 2 space left parenthesis 2 straight a space plus space 4 straight d space right parenthesis space plus 7 over 2 space left parenthesis 2 straight a space plus space 6 straight d right parenthesis space equals space 167
rightwards double arrow space 5 space left parenthesis straight a space plus space 2 straight d right parenthesis space space plus 7 space left parenthesis straight a space plus 3 straight d right parenthesis space equals 167

rightwards double arrow 12 straight a space space plus space 31 straight d space equals space 167 space space.. left parenthesis straight i right parenthesis

Also comma

straight S subscript 10 space equals space 235
rightwards double arrow space 10 over 2 space left parenthesis 2 straight a space plus 9 straight d right parenthesis space equals space 23

rightwards double arrow 2 straight a space plus 9 straight d space equals space 47 space space... space left parenthesis ii right parenthesis

solving space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space ge
straight a equals space 1 space and space straight d equals 5
Hence comma space required space AP space is space 1 comma space 6 comma space 11 space space

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