# Math:A.P.

The 1st,3rd and 9th terms of an A.P are the three terms of a G.P.if the 7th term of the A.P is 19.calculate the 20th term of the A.P and the sum of the first 12terms of the G.P(LEAVING YOUR ANSWER IN INDEX FORM)

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1. for AP:
term1 = a
term3 = a+2d
term9 = a+8d
but term7 = 19 = a+6d
so a = 19-6d

so re-defining:
term1 = 19-6d
term3 = 19-6d+2d = 19-4d
term9 = 19-6d + 8d = 19+2d

these 3 are supposed to be a GP, so
(19-4d)/(19-6d)= (19+2d)/(19-4d)
361 - 152d + 16d^2 = 361 -76d - 12d^2
28d^2 - 76d = 0
d(28d - 76) = 0
d = 0 or d = 76/28 = 19/7

case1 (trivial case) , d = 0
then all terms in the AP would be 19
i.e. 19 19 19 19 ...
of course the first 3 would be a GP also , etc

case 2:
d = 19/7
a = 19 - 6d = 19-6(19/7) = 19/7

for AP, term 20 = a+19d = 19/7+19(19/7) = 380/7

now term1 of AP = term1 of GP
a of GP = 19/7
term 3 of AP = term2 of GP
term2 of GP = a+2d = 19/7 + 2(19/7) = 57/7
so r of GP = (57/7) ÷ (19/7) = 3

sum 12 of GP = a(r^12 - 1)/r
= (19/7)(3^12 - 1)/2

check my arithmetic

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2. the 4th & 9th term of the A.P are 1 & 19 respectively. calculate the 1st term

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3. Solve this problem

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4. Please tell this with clear explanations

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5. Ssss

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6. im so

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7. Right 👍👍

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