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Ap whose third term is 16 and the 7th term exceed the 5 term by 12 |
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Answer» e the First term, a 3 be the third term, a 5 be the 5TH term and a 7 be the 7th termWe know, a n =a+(n−1)d⇒ a 3 =16 [ Given ]⇒ a+(3−1)d=16⇒ a+2d=16 ....... (1)⇒ Now, a 7 −a 5 =12 [ Given ]⇒ [a+(7−1)d]−[a+(5−1)d]=12⇒ 2d=12∴ d=6From equation (1), we geta+2(6)=16a+12=16∴ a=4So first term is 4 We can FIND AP by adding d continuously.∴ Required AP is 4,10,16,22,28,34,40ok my friend |
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