Question

the product of 30th,40th, and 50th terms of a GS is 512. If the 20th terms of an AS is equal to the 40th term of the GS,find the sum of first 39 terms of the AS.

Answer 1

Step-by-step explanation:

Let the 3 number in the geometric sequence be a/r, a and ar.

The product of the three numbers is a^3 = 512, So a = 8.

Now (a/r) +8, a+6, ar form an AP.

So we have: ar - (a+6) = a+6 - (a/r)-8, or

ar^2-ar-6r = ar+6r-a-8r, or

8r^2–8r-6r = 8r+6r-8–8r, or

8r^2–14r-6r +8 = 0

8r^2–20r+8 = 0, or

2r^2–5r +2= 0, or

(2r-1)(r-2) = 0

Hence r =2 or 1/2.

So the three terms of the GP are 4, 8 and 16 or 16, 8 and 4.

Check: 4+8, 8+6, 16 = 12, 14 16, all in AP. Correct.

16+8, 8+6, 4 = 24, 14, 4, all in AP. Correct.

i hope it's helpful for you

Answer 2

Answer: 312

Step-by-step explanation:

Given,

• the product of 30th,40th, and 50th terms of a Geometric Progression (G.P.) is 512.
• The 20th terms of a certain Arithmetic Progression (A.P.) is equal to the 40th term of the G.P.

To Find: The Sum of the first 39 terms of the A.P.

Explanation:

• To solve this question, we need to know about "Arithmetic Progressions" and "Geometric Progressions".
• Now comes the question, “What is an Arithmetic Progression?”
• The answer comes as: An Arithmetic Progression (A.P.) is an arithmetic sequence (of numbers) in which the difference between two consecutive terms is always constant.
• For Example, 2,4,6,8,10 are in Arithmetic Progression.
• The First term of an A.P. is typically denoted by “a”.
• While the Last term of an A.P. is denoted by “l”.
• The common difference of an A.P. is the constant difference between two consecutive terms (as mentioned earlier in the definition of arithmetic progression) and is typically denoted by “d”.
• To find a certain nth term of an A.P. (Tn), we need to use the formula of $Tn=[a+(n-1)d]$.
• And to find the sum of the first n terms of an A.P. (Sn), we need to use the formula of $Sn=\frac{n}{2}[2a+(n-1)d]$
• Then comes the question, “What is a Geometric Progression?”
• The answer comes as: A Geometric Progression (G.P.) is a sequence (of numbers) in which the ratio of two consecutive terms is always constant.
• For Example, 1,2,4,8,16,32 and so on are in Geometric Progression.
• Let us suppose, the First term of a G.P. be denoted by “p”.
• And the Last term of the G.P. be denoted by “q”.
• The common ratio of  the G.P. is the constant ratio of every two consecutive terms and is typically denoted by “r”.
• So to find a certain xth term of an G.P. (Tx), we need to use the formula of $Tx=[p*(r)^{x-1}]$
• And to find the sum of the first x terms of a G.P. (Sx), we need to use the formula of $Sx=[p*(\frac{r^x-1}{r-1})]$
• Now using the above mentioned formulae, let us calculate the 30th,40th, and 50th terms of a Geometric Progression (G.P.).
• Since no data is provided, we will go ahead with our already assumed values.
• 30th Term of a G.P.

         $=[p*(r)^{30-1}]\\=[p*(r)^{29}]\\=pr^{29}$

• Similarly, 40th Term of the G.P. = $pr^{40-1}=pr^{39}$
• And 50th term of the G.P. = $pr^{50-1}=pr^{49}$
• As per condition in the question statement,

        (30th Term of the G.P.) * (40th Term of the G.P.) * (50th Term of the G.P.) = 512

         $or, (pr^{29})*(pr^{39})*(pr^{49})=512\\or, (p^3)*[(r)^{29+39+49}]=512\\or, (p^3)*(r)^{117}=512\\or, (p^3)*(r)^{3*39}=512\\or, (p^3)*(r^{39})^{3}=512\\or, (pr^{39})^{3}=512\\or, \sqrt[3]{(pr^{39})^{3}} = \sqrt[3]{512} \\or,(pr^{39})=8$

• As per the second condition mentioned in the question statement,

       ( $T_{20}$ of an A.P.) = ( $T_{40}$ of a G.P.)

       $or, [a + (20-1)d] = [p*(r)^{40-1}]\\or, [a + 19d]=[pr^{39}]\\or, [a + 19d]=8$

• Therefore, Sum of the first 39 terms of the A.P. using the formula of sum of an A.P. mentioned above

        $S_{39} =\frac{39}{2} [(2*a)+(39-1)d]\\or, S_{39} =\frac{39}{2} [2a+38d]\\or, S_{39} =\frac{39}{2}*2[a+19d]\\or, S_{39} =39*[a+19d]\\or, S_{39} =39*8\\or, S_{39} =312$

The first term of an arithmetic progression is 3000 and the tenth term is 1200. (i) Find the sum of the first 20 terms of the progression. (ii) After how many terms does the sum of the progression become negative?​

brainly.in/question/19707772

If 9th term of an arithmetic progression is 19 the sum of 4th and 7th term is 24, then find Arithmetic progression.

brainly.in/question/36223374

#SPJ2​