Question

find the 11th term from the last term of the AP 10, 7, 4 ..... - 62​

Answer 1

HELLO DEAR,

GIVEN:-

AP is , 10 , 7 , 4,....... , -62

here, a = 10

d = 7 - 10 = -3

l = -62

we know:- l = a + ( n - 1 ) d

-62 = 10 + ( n - 1 ) ( - 3 )

= > -72 = -3( n - 1 )

= > ( n - 1 ) = 72/3

= > n - 1 = 24

= > n = 25

so, last term = 25th

second last term = (25 - 1) = 24th

third last term = ( 25 - 2) = 23th

so, on

also, nth term from the last is ( n - r + 1) th

now we have to find 11th term from last term = (25 + 1 - 11)th = 15th

so, we have to find (a_15).

(a_15) = 10 + (15 - 1)(-3)

= > (a_15) = 10 - 42

= > (a_15) = -32

so, 11th term from the last = 15th from the first = -32

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

$\underline{ \sf \fcolorbox{red}{pink}{ \huge{Solution :)}}}$

Given ,

First term (a) = 10

Common difference (d) = 7 - 10 = -3

Last term (l) = - 62

We know that , the nth term of an AP from end is given by

$\large \sf \fbox{ a_{n} = l - (n - 1)d}$

Substitute the known values , we get

$\sf \mapsto a_{11} = - 62 - (11 - 1)( - 3) \\ \\ \sf \mapsto a_{11} = - 62 - (10)( - 3) \\ \\ \sf \mapsto a_{11} = - 62 + 30 \\ \\ \sf \mapsto a_{11} = - 32$

Hence , the 11th term of an from the last term is -32

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