(A) Since $a, 7, b, 23, c$ are in $AP$,the common difference $d$ is constant.$d = 7 - a = b - 7 = 23 - b = c - 23$.From $b - 7 = 23 - b$,we get $2b =

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(A) Since $a, 7, b, 23, c$ are in $AP$,the common difference $d$ is constant.$d = 7 - a = b - 7 = 23 - b = c - 23$.From $b - 7 = 23 - b$,we get $2b = 30$,so $b = 15$.Using $7 - a = b - 7$,substitute $b = 15$:$7 - a = 15 - 7 = 8$,which gives $a = 7 - 8 = -1$.Using $c - 23 = 23 - b$,substitute $b = 15$:$c - 23 = 23 - 15 = 8$,which gives $c = 8 + 23 = 31$.Thus,$a = -1, b = 15, c = 31$.

Class 10 Mathematics Arithmetic Progressions Mix Examples - Arithmetic Progressions

Medium

Find $a, b$ and $c$ such that the following numbers are in $AP : a, 7, b, 23, c$.

A
$-1, 15, 31$
B
$-1, 14, 30$
C
$-2, 16, 30$
D
$-1, 11, 31$

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Mix Examples - Arithmetic Progressions

For a given $A.P.$,the first term is $-4$ and the common difference is $-5$. Then,the $12^{th}$ term of the $A.P.$ is $\ldots \ldots \ldots$.

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Find $a, b$ and $c$ such that the following numbers are in $AP : a, 7, b, 23, c$.

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For a given $A.P.$,the first term is $-4$ and the common difference is $-5$. Then,the $12^{th}$ term of the $A.P.$ is $\ldots \ldots \ldots$.

The $9^{th}$ term of an $A.P.$ is $0.$ Show that the $29^{th}$ term of the $A.P.$ is two times its $19^{th}$ term.

Verify that each of the following is an $AP$,and then write its next three terms.$a, 2a+1, 3a+2, 4a+3, \ldots$

For an $A.P.$,the $7^{th}$ term is $-1$ and the $16^{th}$ term is $17$. Find the general term of the $A.P.$

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 10 - 6n$. Find the sum of the first $n$ terms of the $A.P.$

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Verify that each of the following is an $AP$,and then write its next three terms.
$a, 2a+1, 3a+2, 4a+3, \ldots$