(C) Given that $k+1, 8, k+9$ are three consecutive terms of an $A.P.$Since the common difference $d$ is constant in an $A.P.$,we have:$8 - (k+1) = (k+

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(C) Given that $k+1, 8, k+9$ are three consecutive terms of an $A.P.$Since the common difference $d$ is constant in an $A.P.$,we have:$8 - (k+1) = (k+9) - 8$$8 - k - 1 = k + 1$$7 - k = k + 1$$7 - 1 = k + k$$6 = 2k$$k = 3$

Class 10 Mathematics Arithmetic Progressions Mix Examples - Arithmetic Progressions

Medium

If $k+1, 8, k+9$ are three consecutive terms of an $A.P.$,then $k = \ldots$

A
$1$
B
$2$
C
$3$
D
$4$

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Class 10

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Arithmetic Progressions

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Mix Examples - Arithmetic Progressions

For any $A.P.$,$T_{30} - T_{20} = \ldots \ldots \ldots \ldots$

Medium

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The sum of the first three terms of an $AP$ is $33$. If the product of the first and the third term exceeds the second term by $29$,find the $AP$.

Medium

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For an $A.P.$,the first term is $10$ and the $10^{th}$ term is $100$. Then,the sum of the first $10$ terms is..........

Easy

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Find the $10^{th}$ term of the $A.P.$ $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$ (in $\sqrt{2}$)

Medium

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For the $AP: -3, -7, -11, \ldots,$ can we find $a_{30} - a_{20}$ directly without actually finding $a_{30}$ and $a_{20}$? Give reasons for your answer.

Medium

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If $k+1, 8, k+9$ are three consecutive terms of an $A.P.$,then $k = \ldots$

Similar Questions

For any $A.P.$,$T_{30} - T_{20} = \ldots \ldots \ldots \ldots$

The sum of the first three terms of an $AP$ is $33$. If the product of the first and the third term exceeds the second term by $29$,find the $AP$.

For an $A.P.$,the first term is $10$ and the $10^{th}$ term is $100$. Then,the sum of the first $10$ terms is..........

Find the $10^{th}$ term of the $A.P.$ $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$ (in $\sqrt{2}$)

For the $AP: -3, -7, -11, \ldots,$ can we find $a_{30} - a_{20}$ directly without actually finding $a_{30}$ and $a_{20}$? Give reasons for your answer.

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