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)

To find the solution, let's break down the problem step by step:

Step 1: Finding the common difference of the arithmetic progression (A.P)
Given that the 7th term of the A.P is 19, we can find the common difference, denoted as d, using the formula:
a + (n-1)d = term_n
where a is the first term, n is the position of the term, and term_n is the value of the nth term.

Plugging in the given values:
a + (7-1)d = 19
a + 6d = 19

Step 2: Finding the first term of the arithmetic progression (A.P)
We know that the 1st, 3rd, and 9th terms of the A.P form a geometric progression (G.P). Therefore, we can use this information to find the first term of the A.P.

Let the 1st term of the A.P be a, and let the common ratio of the G.P be r.
The 1st, 3rd, and 9th terms of the A.P can be written as:
a, ar^2, and ar^8, respectively.

Since these terms form a G.P, we can write:
(ar^2)^2 = a * ar^8
ar^4 = ar^9

Dividing both sides by ar^4, we get:
r^5 = r^8
r^(8-5) = 1
r^3 = 1

Since r^3 = 1, r can be either 1, e^(i2π/3), or e^(-i2π/3). We will only consider r = 1 for this problem.

Therefore, ar^4 = ar^9 leads to a = a.

Based on this information, we can conclude that a must be equal to zero.

Step 3: Calculating the common difference (d)
Using the equation a + 6d = 19 from Step 1 and substituting a = 0, we can solve for d:
0 + 6d = 19
6d = 19
d = 19/6

Step 4: Calculating the 20th term of the arithmetic progression (A.P)
Using the formula for the nth term of an A.P:
term_n = a + (n-1)d

Substituting a = 0 and d = 19/6, we can find the 20th term:
term_20 = 0 + (20-1)(19/6)
term_20 = 0 + 19(19/6)
term_20 = 19(19/6)

Step 5: Calculating the sum of the first 12 terms of the geometric progression (G.P)
Since we know that the 1st, 3rd, and 9th terms of the A.P form a G.P, we can use these values to find the sum of the first 12 terms of the G.P.

The sum of the first 12 terms of a G.P can be calculated using the formula:
S_n = a * (r^n - 1) / (r - 1)

Plugging in the values:
a = 0 (from Step 2)
r = 1 (from Step 2)
n = 12

S_12 = 0 * (1^12 - 1) / (1 - 1)
S_12 = 0

Therefore, the 20th term of the A.P is 19(19/6), and the sum of the first 12 terms of the G.P is 0.

im so stupid

Right 👍👍

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

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

Solve this problem

Please tell this with clear explanations