The modulus of 11th term from the last of the AP series $10, 7, 4, . . ., - 62$ is____

Open in App

Solution

$10, 7, 4, . . . . . . . . . . . . . . - 62$
$F o r m u l a u s e d$
$T_{n} = a + (n - 1) d$
$10, 7, 4......... - 62$
$a = 10$
$d = 7 - 10 = - 3$
$T_{n} = - 62$
$T_{n} = a + (n - 1) d$
$- 62 = 10 + (n - 1) (- 3)$
$\Rightarrow - 62 = 10 + 3 - 3 n$
$\Rightarrow - 62 - 13 = - 3 n$
$n = 25$
$n t h f r o m {last=n-T}_{n} + 1 t e r m f r o m b e g i n i n g$
$25 - 11 + 1 = 15^{t h} t e r m f r o m b e g i n i n g$
$T_{15} = 10 + (15 - 1) (- 3)$
$= 10 + (14) (- 3)$
$= 10 - 42$
$= - 32$

Suggest Corrections

Similar questions

11 t h

A P : 10, 7, 4, . . ., - 62

10, 7, 4, . . . . ., - 62

11^{t h}

11^{t h}

A . P

10, 7, 4, \dots \dots . - 62

Find the 11th term of the last term ( towards the first term) of the arithmetic progression 10, 7, 4, .......-62

Join BYJU'S Learning Program