(C) The sum of $n$ terms of an $AP$ is given by the formula: $S_{n} = \frac{n}{2}(a + a_{n})$.Given $a = 1$,$a_{n} = 20$,and $S_{n} = 399$.Substitutin

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(C) The sum of $n$ terms of an $AP$ is given by the formula: $S_{n} = \frac{n}{2}(a + a_{n})$.Given $a = 1$,$a_{n} = 20$,and $S_{n} = 399$.Substituting these values into the formula:$399 = \frac{n}{2}(1 + 20)$$399 = \frac{n}{2}(21)$$399 = 10.5n$$n = \frac{399}{10.5}$$n = 38$.Thus,the value of $n$ is $38$.

Class 10 Mathematics Arithmetic Progressions Mix Examples - Arithmetic Progressions

Medium

In an $AP$,if $a = 1$,$a_{n} = 20$,and $S_{n} = 399$,then $n$ is:

A
$19$
B
$21$
C
$38$
D
$42$

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Arithmetic Progressions

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Mix Examples - Arithmetic Progressions

If the $9^{\text{th}}$ term of an $AP$ is zero,prove that its $29^{\text{th}}$ term is twice its $19^{\text{th}}$ term.

Medium

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Two distinct terms of an $A.P.$ cannot be........

Easy

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The $n^{th}$ term of an $A.P.$ is given by $T_n = 5n - 2$. Then,the $12^{th}$ term of the $A.P.$ is:

Easy

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Show that the sum of an $AP$ whose first term is $a,$ the second term is $b,$ and the last term is $c,$ is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}.$

Difficult

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The $n^{th}$ term of an $A.P.$ is given by $T_{n} = -4n + 15$. Find the $15^{th}$ term and the common difference of the $A.P.$

Medium

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In an $AP$,if $a = 1$,$a_{n} = 20$,and $S_{n} = 399$,then $n$ is:

Similar Questions

If the $9^{\text{th}}$ term of an $AP$ is zero,prove that its $29^{\text{th}}$ term is twice its $19^{\text{th}}$ term.

Two distinct terms of an $A.P.$ cannot be........

The $n^{th}$ term of an $A.P.$ is given by $T_n = 5n - 2$. Then,the $12^{th}$ term of the $A.P.$ is:

Show that the sum of an $AP$ whose first term is $a,$ the second term is $b,$ and the last term is $c,$ is equal to $\frac{(a+c)(b+c-2a)}{2(b-a)}.$

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = -4n + 15$. Find the $15^{th}$ term and the common difference of the $A.P.$

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