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Show that a1,a2,form an Ap where an is defined :an=3+4n |
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Answer» i) a n =3+4n given a 1 =3+4(1)=7 a 2 =3+4(2)=11 a 2 =3+4(3)=15 ∴d=a 3 −a 1 =11−7=4 Here a=7, d=4 and n=15 By USINGS n = 2 n [2a+(n−1)d] we have, S 15 = 2 15 [2×7+(15−1)4] = 2 15 (14+56) = 2 15 ×70=525 (ii) a n =9−5n given a 1 =9−5(1)=9−5=4 a 2 =9−5(2)=9−10=−1 a 3 =9−5(3)=9−15=−6 ∴d=a 2 −a 1 =−6−(−1)=−5 Here a=4,d=−5 and n=15 By using S n = 2 n [2a+(n−1)d] we have, S 15 = 2 15 [2×4+(15−1)(−5)] = 2 15 (8−70) = 2 15 ×(−62)=−465 |
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