Question

The 10th yerm of AP is 52 and 6th term is 82. Find the32nd term and general form of (An)

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{32nd\:term=-113}}}$

$\green{\tt{\therefore{A_{n}=127-7.5n}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Fiven:}} \\ \tt: \implies 10th \: of \: A.P = 52 \\ \\ \tt: \implies 6th \: of \: A.P = 82 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies 32nd \: term \: of \: A.P= ? \\ \\ \tt: \implies nth\: term \: of \: A.P = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies a_{10} = 52 \\ \\ \tt: \implies a + 9d = 52 \\ \\ \tt: \implies a = 52 - 9d - - - - - (1) \\ \\ \bold{Similarly : } \\ \tt: \implies a_{6} = 82 \\ \\ \tt: \implies a + 5d = 82 - - - - - (2) \\ \\ \text{Putting \: value \: of \: a \: in \: (2)} \\ \tt: \implies 52 - 9d+5d = 82 \\ \\ \tt: \implies 52 - 82 = 4d\\ \\ \tt: \implies d = -\frac{30}{4} \\ \\ \green{\tt: \implies d = -\frac{15}{2}} \\ \\ \text{Putting \: value \: of \: d \: in \: (1)} \\ \tt: \implies a = 52 - 9 \times -7.5 \\ \\ \tt: \implies a = 52 +67.5\\ \\ \green{\tt: \implies a = 119.5} \\ \\ \bold{for \: 32nd \: term : } \\ \tt: \implies a_{n} = a + (n - 1)d \\ \\\tt: \implies a_{32} = 119.5 + (32 - 1) \times -7.5 \\ \\ \tt: \implies a_{32} = 119.5 + 31 \times -7.5 \\ \\ \green{\tt: \implies a_{32} = -113} \\ \\ \bold{For \: nth \: term : } \\ \tt: \implies a_{n} =119.5+ (n - 1) \times -7.5 \\ \\ \tt: \implies a_{n} = 119.5- 7.5n+7.5 \\ \\ \green{\tt: \implies a_{n} = 127-7.5n}$

Answer 2

..

$\tt{\huge{\purple{ ................ }}}$

QUESTION -

The 10th yerm of AP is 52 and 6th term is 82.

Find the32nd term and general form of (An).

SOLUTION -

From the above Question, we have the following information....

The 10th yerm of AP is 52 and 6th term is 82.

FORMULAE USED -

We know that a general term of an ap, An can be expressed as :

$\sf{\green{ \leadsto{ \boxed{ \boxed { A_{n} = a + ( n -1 ) d }}}}}$

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Now let us use the above general formula to express the given terms.

The 10th term of An AP refers to the fact that the value of n = 10

The 32nd  term of An AP refers to the fact that the value of n = 32

Similarly in the sixth term ,  n = 6.

So, representing them in the general Fromulae :

$\sf{\orange{ \leadsto{ \boxed{ \boxed { A_{10} = a + ( 9 ) d = 52 }}}}} .......... .... [ 1 ]$

$\sf{\blue{ \leadsto{ \boxed{ \boxed { A_{6} = a + ( 5 ) d = 82}}}}} .......... .... [ 2 ]$

$\sf{\pink{ \leadsto{ \boxed{ \boxed { A_{32} = a + ( 32 ) d = ? }}}}} .......... .... [ 3 ]$

Now Let Us Subtract Equation 1 from Equation 2.

So, .

$\sf{\red{ \leadsto{ \boxed{ \boxed { 4d = -30 => d = -7.5}}}}}$

Placing the above obtained value of D in either of the equations 1 or 2, we get a = -119.5

Now,

Let us Substute these values in Equation 3

$\sf{\pink{ \leadsto{ \boxed{ \boxed { A_{32} = -119.5 - 31 \times 7.5 = -113}}}}} .......... .... [ A ]$

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