Question

10. The sum of the first ten terms of an arithmetic progression is equal to 155 and the sum of the first two terms of a geometric sequence is 9, of the first term of arithmetic sequence is equal to the common ratio of geometric sequence and the first term of geometric sequence is equal to the common difference of arithmetic progression. Find some terms of arithmetic progression and some terms of geometric progression. 99​

Answer 1

Hi,

Explanation:

Let's assume:

`a_1\ the\ first\ term\ of\ the\ AP\\r\ the\ common\ difference\ of\ the\ AP\\a_1,a_2=a_1+r,a_3=a_1+2r,...,a_(10)=a_1+9r\\\displaystyle \sum_(i=1)^(10)\ a_i=10*a_1+45*r=155\\\\`

`b_1\ the\ first\ term\ of\ the\ GP\\r'\ the\ ratio \ of\ the\ GP\\b_1+b_2=b_1+b_1*r'=b_1*(1+r')=9\\`

`\left\{\begin{array}{ccc}10a_1+45r&=&155\\b_1(1+r')&=&9\\a_1&=&r'\\b_1&=&r\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}r'&=&(155-45r)/(10)\\r+r*r'&=&9\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}r'&=&(31-9r)/(2)\\r+r*((31-9r)/(2))&=&9\\\end {array}\right.\\\\9r^2-33r+18=0\\\\\Delta=33^2-4*9*18=441=21^2\\\\r=(33+21)/(18) =3\ or\ r=(33-21)/(18) =(2)/(3) \\`

`1)\\r=3\\b_1=3\\r'=(31-9*3)/(2) =2\\a_1=2\\\\a_n=2+3*(n-1)\\2,5,8,11,14,...\\b_n=3*2^(n-1)\\3,6,12,24,48,...\\`

2)

`r=b_1=(2)/(3) \\r'=a_1=(25)/(2) \\\\a_n=(25)/(2) +(2)/(3) ^(n-1)\\\\b_n=(2)/(3) *{((25)/(2))}^(n-1)`