Let a n be the n t h term of the series 5 + 8 + 14 + 23 + 35 + 50 + ... and S n = k = 1 n a k . Then S 30 - a 40 is equal to [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $a_{n}$ be the $n^{t h}$ term of the series 5 + 8 + 14 + 23 + 35 + 50 + ... and $S_{n} = \sum_{k = 1}^{n} a_{k} .$ Then $S_{30} - a_{40}$ is equal to [2023]

1 11280

2 11290

3 11310

4 11260

Ans.
(2)

Let $S = 5 + 8 + 14 + 23 + 35 + 50 + \dots + a_{n}$

$S = 5 + 8 + 14 + 23 + \dots + a_{n - 1} + a_{n}$

Subtracting the above two equations, we get

$0 = 5 + 3 + 6 + 9 + \dots up to (n - 1) terms - a_{n}$

$\Rightarrow a_{n} = 5 + (\frac{n - 1}{2}) [6 + (n - 2) \times 3]$

$= 5 + \frac{3 n (n - 1)}{2} = \frac{1}{2} (3 n^{2} - 3 n + 10)$

Now, $S_{n} = \sum_{k = 1}^{n} a_{k} = \frac{1}{2} \sum_{k = 1}^{n} (3 n^{2} - 3 n + 10)$

   $= \frac{1}{2} [\frac{3 n (n + 1) (2 n + 1)}{6} - \frac{3 n (n + 1)}{2} + 10 n]$

   $= \frac{1}{2} [\frac{n (n + 1) (2 n + 1)}{2} - \frac{3 n (n + 1)}{2} + \frac{20 n}{2}]$

$= \frac{1}{2} [\frac{2 n^{3} + 3 n^{2} + n - 3 n^{2} - 3 n + 20 n}{2}]$

   $= \frac{1}{2} [\frac{2 n^{3} + 18 n}{2}] = \frac{1}{2} n (n^{2} + 9)$

Now, $S_{30} - a_{40} = \frac{1}{2} \times 30 (30^{2} + 9) - \frac{1}{2} (3 \times 40^{2} - 3 \times 40 + 10)$

   $= 13635 - 2345 = 11290$

Want Us to Email you About Special Offers & Updates?