Question

arithmetic progression. SOLVE:

Answer 1

Given :-

◉ a3 = 15

◉ S10 = 125

To Find :-

◉ d

◉ a10

Solution :-

We know,

⇒ Third term = a + 2d

∴ a + 2d = 15 ...(1) {given}

Now,

⇒ Sum of first 10 terms = 125 {given}

⇒ 10/2 [ 2a + 9d ] = 125

⇒ 5( 2a + 9d ) = 125

⇒ 10a + 45d = 125 ...(2)

Multiply (1) by 10, We get

⇒ 10a + 20d = 150 ...(3)

Subtract (2) from (3), we get

⇒ 10a + 20d - 10a - 45d = 150 - 125

⇒ -25d = 25

⇒ d = -1 ...(4)

Now, Put d = -1 in (1)

⇒ a + 2×-1 = 15

⇒ a - 2 = 15

⇒ a = 17 ...(5)

So,

⇒ 10th term = a + 9d

⇒ 10th term = 17 + 9×-1 { from (4) & (5) }

⇒ 10th term = 17 - 9

⇒ 10th term = 8

Hence, a10 = 8 and d = -1

Formulae related to this topic:

$\star \sf \qquad S_{n} = \dfrac{n}{2} \{ 2a + (n - 1)d \} \\ \\ \star \sf \qquad a_{n} = a + (n - 1)d$

Answer 2

ᴀɴsᴡᴇʀ

ɢɪᴠᴇɴ:— $a_3 = 15$

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ:—

$a_3 = a + 2d$

$15 = a + 2d$

$a = 15 - 2d$ ________(ɪ)

ɢɪᴠᴇɴ :— $s_{10} = 125$

ᴡᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ:—

$S_{10} = \frac{10}{2} [2a + (10 - 1)d]$

$125 = 5(2a + 9d)$

$\frac{125}{5} = 2a + 9d$

$25 = 2a + 9d$

$25 = 2(15 - 2d) + 9d$

$25 = 30 - 4d + 9d$

$25 - 30 = 5d$

$- 5 = 5d$

$d = 1$

ᴛʜᴇʀᴇғᴏʀᴇ:—

d = 1

ᴘᴜᴛᴛɪɴɢ d = 1 ɪɴ (ɪ):-

$a = 15 - (2 \times ( - 1))$

$a = 15 + 2$

$a = 17$

$a_{10} = a + 9d$

$a_{10} = 17 + 9 \times ( - 1)$

$a_{10} = 17 - 9$

$a_{10} = 8$

ʜᴇɴᴄᴇ:—

a10 = 8