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Question Number 210630 by peter frank last updated on 14/Aug/24

Answered by mm1342 last updated on 14/Aug/24

a_n =a+(n−1)d & b_n =2^(n−1) g c_n =a_n +b_n & c_1 =57 & c_3 =171 & c_5 =645 ⇒a+g=57 & a+2d+4g=171 & a+4d+16d=645 ⇒a=17 & d=−3 & g=40 ⇒c_n =40×2^(n−1) −3n+20 ✓

an=a+(n−1)d&bn=2n−1gcn=an+bn&c1=57&c3=171&c5=645⇒a+g=57&a+2d+4g=171&a+4d+16d=645⇒a=17&d=−3&g=40⇒cn=40×2n−1−3n+20✓

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