Find The First Three Terms Of The Geometric Sequence Defined By .

Find the 10th term in the sequence whose first three terms are 3, 3 and 6 and each term after the second term is the sum of the two terms proceeding it?

This is a very interesting sequenceLook at the pattern 3,3,6,9,15,24………..We usually see that in arithmetic progression there is always a common difference, say dbut here when we subtract the previous number from the successive number we see3–3=06–3=39–6=315–9=624–15–9so the differences are like this 0,3,3,6,9……. the same sequence which was in the original sequenceNow here is something more interesting i.e.3,3,6,9,15,24….. is equivalent to 3(1,1,2,3,5,8…..)As we know 1,1,2,3,5,8 ….. is a Fibonacci sequenceAnd this sequence is 3 time of the Fibonacci sequenceWhenever we are dividing the successive number by it’s preceding number we always get an approx of 1.618Ex: 21/13=1.61513/8=1.6258/5=1.65/3=1.66666This is golden Ratiodenoted by ϕ (phi) ϕ = 1/2+√5/2 ≈ 1.618033988 (the golden ratio)Now to find the nth term of this Fibonacci sequence let us use this formulaThe terms of the Fibonacci sequence can be defined as:X1=1X2=1Xn+2= X+Xn+1The general term can also be expressed by a formula:Xn ≈ ϕ^n / √5= (1.618^10) / 2.2360679775=54.992083rounding up 54.992083 we get 55.But the original sequence was 3,3,6,9,15,24….which is equivalent to 3 (1,1,2,3,5,8…..)So Here 10th term of the sequence will be 3 * 55 = 165Hope this could help you.Thank youSweta SharanCoordinator at Indradhanush Academyindradhanush.guru

How do I find the first term of a geometric sequence when given the number of terms, ratio, and sum?

Ans : a = 0.0156Solution :Here ,S = 21/2048r = -1/2 = -0.5n = 6● Consider the 2nd formula21/2048 = a (1 - (-0.5)^6) /(1 -(-0.5))》apply the powers and calculate the values21/2048 = a (1 - (0.0156) /(1 + 0.5)》simple calculations21/2048 = a (0.984)/(1.5)》switching the known and un known terms(21/2048) × (1.5/0.984) = a》 calculate the value of a.☆ a = 0.0156Therefore the solution is complete

The first term of a geometric sequence is 16 and the fifth term is 9. What is the value of the seventh term?

a1=16 , a5 =9 , a7=?a5=a1(r)*n-19=16 r*4 , r*4 =0.5625 ,to get the value of use 4th square root both side of the equation , then you will get:-r=0.866025403after that 7 th term will be as the following:a7= 16(0.866025403)*6 = 6.75 or as fraction 27/4.therefore , the 7th term of G.P is 6.75 or as fraction 27/4.

For a real number t, the first three terms of an arithmetic sequence are 3t – 4, 2t + 3, and 6t – 2. What is the numerical value of the fourth term?

We know that in an Arithmetic Sequence the difference between one term and the next is a constant.Now, having this in mind, we see that there are three terms. (3t-4; 2t+3; 6t-2).By combining these two facts we conclude that 6t-2-(2t+3) = 2t+3-(3t-4).After solving this equation we get that t=2.4Now we insert this number into the previuosly mentioned terms:1st term: 3 * 2.4 – 4 = 3.22nd term: 2 * 2.4 + 3 = 7.83rd term: 6 * 2.4 - 2 = 12.4The difference between two consecutive terms is 4.6So, in order to find the 4th number, we simply add 4.6 to the 3rd number: 12.4 + 4.6.And so, the answer is 17.

The first three terms of a sequence are 1, 2 and 3. Each subsequent term is the sum of the three previous terms. What is the 11th term of this sequence? Can you explain the answer so a middle schooler will fully understand?

All this question is asking is that in order to figure out the next unknown value in the sequence, you have to add the previous three. So in order to get the fourth value, just add the three known values before it, 1+2+3=6. Therefore the next known term is 6. The fifth term is the sum of the previous known terms, 2+3+6=11. This means we now know the fifth term is 11. This pattern will continue until you get the 11th value.3+6+11=206+11+20=3711+20+37=6820+37+68=11737+68+117=22268+117+222=40711th number is 407Thanks for giving me a reason to procrastinate from calculus hw

In a geometric series, if the sum of the first 10 terms is 4 and the sum of the 11th to the 30th term is 48, then what is the sum of the 31st to the 60th term?

A variation of the answers just written by Alexander Talalai and Conner Davis. Not substantially different.Let [math]t_n[/math] denote the [math]n[/math] th term of the geometric sequence. If the common ratio is [math]r[/math], then [math]t_n=t_1 \cdot r^{n-1}[/math]. The sum of [math]n[/math] terms of the same sequence, starting with [math]t_m[/math], is [math]t_m \cdot R[/math], where [math]R[/math] equals [math](1-r^n)/(1-r)[/math] when [math]r \ne 1[/math], and [math]n[/math] when [math]r=1[/math].Therefore[math]t_1 \cdot R = 4[/math], [math](t_{11} \cdot R)+(t_{21} \cdot R) = t_1(r^{10}+r^{20}) \cdot R = 48[/math].Substituting the first equation in the second, we get [math]r^{10}+\left(r^{10}\right)^2=12[/math]. This is a quadratic in [math]r^{10}[/math], solving which gives [math]r^{10}=3[/math] (we neglect [math]r^{10}=-4[/math], since that is only possible if we allow non-real values for [math]r[/math]).Hence, the required sum is[math](t_{31} \cdot R) + (t_{41} \cdot R)+(t_{51} \cdot R) = (r^{30}+r^{40}+r^{50}) \cdot t_1 \cdot R [/math][math]= 4(3^3 + 3^4 + 3^5) = 1404[/math]. [math]\blacksquare[/math]