# The sum of 4 consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of the middle terms is 7:15. Find the numbers.

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Hint: We first assume the four consecutive terms of the AP using first term a and common difference d. Then we proceed according to the given situation in the question to obtain a and d.

Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d), where a is the first term and d is the common difference.

So, according to the question, the sum of 4 consecutive numbers in an AP is 32,

Therefore, \[a-3d\text{ }+\text{ }a\text{ }-\text{ }d\text{ }+\text{ }a\text{ }+\text{ }d\text{ }+\text{ }a\text{ }+\text{ }3d\text{ }=\text{ }32\]

\[\Rightarrow 4a\text{ }=\text{ }32\]

\[\Rightarrow a\text{ }=\text{ }\dfrac{32}{4}=8\]……….(i)

Now, given that the ratio of the product of the first and the last term to the product of the middle terms is 7:15

\[\Rightarrow \dfrac{\left( a\text{ }-\text{ }3d \right)\left( a\text{ }+\text{ }3d \right)}{\left( a\text{ }-\text{ }d \right)\left( a\text{ }+\text{ }d \right)}\text{ }=\text{ }\dfrac{7}{15}\]

Cross multiplying above equation we have,

$ \Rightarrow 15\left( a{}^\text{2}\text{ }-\text{ }9d{}^\text{2} \right)\text{ }=\text{ }7\left( a{}^\text{2}\text{ }-\text{ }d{}^\text{2} \right) \\ $

$ \Rightarrow 15a{}^\text{2}\text{ }-\text{ }135d{}^\text{2}\text{ }=\text{ }7a{}^\text{2}\text{ }-\text{ }7d{}^\text{2} \\ $

$ \Rightarrow 15a{}^\text{2}\text{ }-\text{ }7a{}^\text{2}\text{ }=\text{ }135d{}^\text{2}\text{ }-\text{ }7d{}^\text{2} \\ $

$ \Rightarrow 8a{}^\text{2}\text{ }=\text{ }128d{}^\text{2} \\ $

Putting the value of a = 8, obtained in equation(i) we get,

$ \Rightarrow 8\left( 8 \right){}^\text{2}\text{ }=\text{ }128d{}^\text{2} \\ $

$ \Rightarrow 128d{}^\text{2}\text{ }=\text{ }512 \\ $

$ \Rightarrow d{}^\text{2}\text{ }=\text{ }\dfrac{512}{128} \\ $

$ \Rightarrow d{}^\text{2}\text{ }=\text{ }4 \\ $

$ \Rightarrow d\text{ }=\text{ }\pm 2 \\ $

Now since the common difference d has two values, therefore we will get two series.

Substituting the values of first term and common difference d = 2 in (a - 3d), (a - d), (a + d) and (a + 3d), we get the four consecutive terms of the AP as \[8\text{ }-\text{ }6\text{ }=\text{ }2\], \[8\text{ }-\text{ }2\text{ }=\text{ }6\], \[8\text{ }+\text{ }2\text{ }=\text{ }10\] and \[8\text{ }+\text{ }6\text{ }=\text{ }14\].

Again, substituting the value of d = -2 in (a - 3d), (a - d), (a + d) and (a + 3d), we get the four consecutive terms of the AP as 8+6 = 14, 8+2 = 10, 8-2 = 6, 8-6 = 2.

Hence, four consecutive numbers of the first AP are 2, 6, 10 and 14.

And the four consecutive terms of second AP are 14, 10, 6, 2.

Note: The possibility of mistake in the question can be at the point where you are assuming the first four terms of the AP. Always remember to take terms with a common difference in between them. Be careful at the time of considering the value of d. Don’t reject any value until there is no condition given to selecting any particular value.

__Complete step-by-step solution -__Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d), where a is the first term and d is the common difference.

So, according to the question, the sum of 4 consecutive numbers in an AP is 32,

Therefore, \[a-3d\text{ }+\text{ }a\text{ }-\text{ }d\text{ }+\text{ }a\text{ }+\text{ }d\text{ }+\text{ }a\text{ }+\text{ }3d\text{ }=\text{ }32\]

\[\Rightarrow 4a\text{ }=\text{ }32\]

\[\Rightarrow a\text{ }=\text{ }\dfrac{32}{4}=8\]……….(i)

Now, given that the ratio of the product of the first and the last term to the product of the middle terms is 7:15

\[\Rightarrow \dfrac{\left( a\text{ }-\text{ }3d \right)\left( a\text{ }+\text{ }3d \right)}{\left( a\text{ }-\text{ }d \right)\left( a\text{ }+\text{ }d \right)}\text{ }=\text{ }\dfrac{7}{15}\]

Cross multiplying above equation we have,

$ \Rightarrow 15\left( a{}^\text{2}\text{ }-\text{ }9d{}^\text{2} \right)\text{ }=\text{ }7\left( a{}^\text{2}\text{ }-\text{ }d{}^\text{2} \right) \\ $

$ \Rightarrow 15a{}^\text{2}\text{ }-\text{ }135d{}^\text{2}\text{ }=\text{ }7a{}^\text{2}\text{ }-\text{ }7d{}^\text{2} \\ $

$ \Rightarrow 15a{}^\text{2}\text{ }-\text{ }7a{}^\text{2}\text{ }=\text{ }135d{}^\text{2}\text{ }-\text{ }7d{}^\text{2} \\ $

$ \Rightarrow 8a{}^\text{2}\text{ }=\text{ }128d{}^\text{2} \\ $

Putting the value of a = 8, obtained in equation(i) we get,

$ \Rightarrow 8\left( 8 \right){}^\text{2}\text{ }=\text{ }128d{}^\text{2} \\ $

$ \Rightarrow 128d{}^\text{2}\text{ }=\text{ }512 \\ $

$ \Rightarrow d{}^\text{2}\text{ }=\text{ }\dfrac{512}{128} \\ $

$ \Rightarrow d{}^\text{2}\text{ }=\text{ }4 \\ $

$ \Rightarrow d\text{ }=\text{ }\pm 2 \\ $

Now since the common difference d has two values, therefore we will get two series.

Substituting the values of first term and common difference d = 2 in (a - 3d), (a - d), (a + d) and (a + 3d), we get the four consecutive terms of the AP as \[8\text{ }-\text{ }6\text{ }=\text{ }2\], \[8\text{ }-\text{ }2\text{ }=\text{ }6\], \[8\text{ }+\text{ }2\text{ }=\text{ }10\] and \[8\text{ }+\text{ }6\text{ }=\text{ }14\].

Again, substituting the value of d = -2 in (a - 3d), (a - d), (a + d) and (a + 3d), we get the four consecutive terms of the AP as 8+6 = 14, 8+2 = 10, 8-2 = 6, 8-6 = 2.

Hence, four consecutive numbers of the first AP are 2, 6, 10 and 14.

And the four consecutive terms of second AP are 14, 10, 6, 2.

Note: The possibility of mistake in the question can be at the point where you are assuming the first four terms of the AP. Always remember to take terms with a common difference in between them. Be careful at the time of considering the value of d. Don’t reject any value until there is no condition given to selecting any particular value.

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