(C) Given that the first term $a = 5$ and the $19^{th}$ term $l = T_{19} = 95$.The number of terms $n = 19$.The formula for the sum of $n$ terms of an

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(C) Given that the first term $a = 5$ and the $19^{th}$ term $l = T_{19} = 95$.The number of terms $n = 19$.The formula for the sum of $n$ terms of an $A.P.$ when the first and last terms are known is $S_{n} = \frac{n}{2}(a + l)$.Substituting the values: $S_{19} = \frac{19}{2}(5 + 95)$.$S_{19} = \frac{19}{2}(100)$.$S_{19} = 19 \times 50 = 950$.Therefore,the sum of the $19$ terms is $950$.

Class 10 Mathematics Arithmetic Progressions Mix Examples - Arithmetic Progressions

Medium

For a given $A.P.$,the first term is $5$ and the $19^{th}$ term is $95$. Then,the sum of its $19$ terms is $\ldots \ldots \ldots \ldots$

A
$1095$
B
$1000$
C
$950$
D
$1900$

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Class 10

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Arithmetic Progressions

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Mix Examples - Arithmetic Progressions

The sum of $n$ terms of an $A.P.$ is given by $S_{n} = 2n^{2} + 5n$. Then,the $n^{th}$ term of the $A.P.$ $T_{n} = \dots$

Medium

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The sum of the first five multiples of $3$ is

Medium

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The ratio of the sums of first $n$ terms of two $A.P.s$ is $(5n - 3) : (7n + 2)$. Find the ratio of their $m^{th}$ terms.

Difficult

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The $n^{th}$ term of an $A.P.$ is given by $T_n = 4n + 3$. Find the sum of the first $60$ terms of the $A.P.$

Easy

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The eighth term of an $AP$ is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{\text{th}}$ term.

Difficult

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For a given $A.P.$,the first term is $5$ and the $19^{th}$ term is $95$. Then,the sum of its $19$ terms is $\ldots \ldots \ldots \ldots$

Similar Questions

The sum of $n$ terms of an $A.P.$ is given by $S_{n} = 2n^{2} + 5n$. Then,the $n^{th}$ term of the $A.P.$ $T_{n} = \dots$

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The ratio of the sums of first $n$ terms of two $A.P.s$ is $(5n - 3) : (7n + 2)$. Find the ratio of their $m^{th}$ terms.

The $n^{th}$ term of an $A.P.$ is given by $T_n = 4n + 3$. Find the sum of the first $60$ terms of the $A.P.$

The eighth term of an $AP$ is half its second term and the eleventh term exceeds one third of its fourth term by $1$. Find the $15^{\text{th}}$ term.

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