The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

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.
The sum of first and eighth terms of an A.P. is 32 and their product is 60. Find the first term and common difference of the A.P. Hence, also find the sum of its first 20 terms.

Ans.
Let a and a8 be first and eight terms of A.P..

Let common difference be d.

$∴ a + a_{8} = 32 (given)$

$\Rightarrow a + [a + (8 - 1) d] = 32 [∵ a_{n} = a + (n - 1) d]$

$\Rightarrow a + (a + 7 d) = 32 \Rightarrow a + 7 d = 32 - a$ ...(i)

$Also, a \cdot a_{8} = 60 (given)$

$\Rightarrow a \cdot [a + (8 - 1) d] = 60 \Rightarrow a (a + 7 d) = 60$

$\Rightarrow a (32 - a) = 60 [from eq (i)]$

$\Rightarrow 32 a - a^{2} = 60 \Rightarrow a^{2} - 32 a + 60 = 0$

$\Rightarrow a^{2} - 30 a - 2 a + 60 = 0 \Rightarrow a (a - 30) - 2 (a - 30) = 0$

$\Rightarrow (a - 30) (a - 2) = 0 \Rightarrow a = 2, 30$

For $a = 2,$ from eq. (i), we get $2 + 7 d = 32 - 2$

$\Rightarrow 7 d = 28 \Rightarrow d = 4$

For $a = 30,$ from eq. (i), we get $30 + 7 d = 32 - 30$

$\Rightarrow 7 d = - 28 \Rightarrow d = - 4$

for $(a, d) = (2, 4)$

$a_{8} = 2 + 7 \times 4 = 30$

$∴ a + a_{8} = 32$ and $a \cdot a_{8} = 60$

for $(a, d) = (30, - 4)$

$a_{8} = 30 + 7 (- 4) = 2$

$∴ a + a_{8} = 32$ and $a \cdot a_{8} = 60$

Taking $(a, d) = (2, 4)$

$S_{20} = \frac{20}{2} [2 \times 2 + (20 - 1) \times 4] [∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]]$

$= 10 [4 + 19 \times 4] = 40 \times 20 = 800$

Taking $(a, d) = (30, - 4)$

$S_{20} = \frac{20}{2} [2 \times 30 + (20 - 1) \times (- 4)] [∵ S_{n} = \frac{n}{2} [2 a + (n - 1) d]]$

$= 10 [60 + 19 (- 4)]$

$= 10 (60 - 76)$

$= 10 \times (- 16) = - 160$

Want Us to Email you About Special Offers & Updates?