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Question: How do you find the next three terms in\(7,14,28,56............\) ?...

How do you find the next three terms in7,14,28,56............7,14,28,56............ ?

Explanation

Solution

There are mainly two types of sequences
Arithmetic sequences: This is a type of sequence where the difference between consecutive terms is always the same.
Geometric sequence: This is a type of sequence where the ratio between consecutive terms is always the same.
Now with the help of the above two definitions we will find the given series and accordingly let us solve the problem.

Complete step by step solution:
Given
7,14,28,56............      ........................(i)7,14,28,56............\;\;\;........................\left( i \right)
Now we can see that the given series is a geometric series and we need to find the next 3 terms of it.
We have the basic definition of geometric sequence as:
Geometric sequence:
This is a type of sequence where the ratio between consecutive terms is always the same and we call that ratio as common ratio.
r=ratio of successive termsr = {\text{ratio of successive terms}}
Also the general term for a geometric sequence is given by:
an=a×rn−1{a_n} = a \times {r^{n - 1}}
Now let’s take the common ratio of the successive terms of the given sequence 7,14,28,56............7,14,28,56............ such that, we can write:
r=147=2 r=2814=2 r=5628=2  r = \dfrac{{14}}{7} = 2 \\\ r = \dfrac{{28}}{{14}} = 2 \\\ r = \dfrac{{56}}{{28}} = 2 \\\
So on observing the above results we can conclude that the common ratio:
r=2r = 2
Now we have been givena1,a2,a3  and  a4{a_1},{a_2},{a_3}\;{\text{and}}\;{a_4}, and we need to finda5,a6  and  a7{a_5},{a_6}\;and\;{a_7}.
For that we have the equationan=a×rn−1{a_n} = a \times {r^{n - 1}}:
Such that we can write:

a5=a×rn−1 =7×25−1 =7×24 =7×16 =112.....................................(ii) a6=a×rn−1 =7×26−1 =7×25 =7×32 =224.....................................(iii) a7=a×rn−1 =7×27−1 =7×26 =7×64 =448.....................................(iv)  {a_5} = a \times {r^{n - 1}} \\\ = 7 \times {2^{5 - 1}} \\\ = 7 \times {2^4} \\\ = 7 \times 16 \\\ = 112.....................................\left( {ii} \right) \\\ {a_6} = a \times {r^{n - 1}} \\\ = 7 \times {2^{6 - 1}} \\\ = 7 \times {2^5} \\\ = 7 \times 32 \\\ = 224.....................................\left( {iii} \right) \\\ {a_7} = a \times {r^{n - 1}} \\\ = 7 \times {2^{7 - 1}} \\\ = 7 \times {2^6} \\\ = 7 \times 64 \\\ = 448.....................................\left( {iv} \right) \\\

Therefore from (ii), (iii) and (iv) we can write the next three terms of the geometric sequence 7,14,28,56............7,14,28,56............ a5,a6  ,a7{a_5},{a_6}\;,{a_7} as a5=112,a6=224,a7=448{a_5} = 112,{a_6} = 224,{a_7} = 448.

Note: General terms of arithmetic and geometric series:
Arithmetic series: an=a+(n−1)d{a_n} = a + \left( {n - 1} \right)d
Geometric series:an=r×an−1{a_n} = r \times {a_{n - 1}}
While doing similar questions one should be very thorough with the properties and formulas regarding the sequences and care must be given to minimize the errors in the calculation process.