Question

if the sum of the first seven terms of an AP is 79 of it 17th terms 289​

Answer 1

Answer:

Given that,

S7 = 49

S17 = 289

S7

= 7/2 [2a + (n - 1)d] S7 = 7/2 [2a + (7 - 1)d]

49 = 7/2 [2a + 16d] =

7 = (a + 3d)

a + 3d = 7 ... (i)

Similarly,

S17 = 17/2 [2a + (17 - 1)d]

289 = 17/2 (2a + 16d)

17 = (a + 8d) = a + 8d = 17.. (ii)

Subtracting equation (i) from equation (i),

5d = 10

d = 2

From equation (i),

a + 3(2) = 7

a + 6 = 7

a = 1

Sn = n/2 [2a + (n - 1)d]

= n/2 [2(1) + (n - 1) x 2]

= n/2 (2 + 2n - 2)

= n/2 (2n)

= n2

Step-by-step explanation:

See the attachment Also...

Answer 2

$\huge\purple{\mathfrak{Answer:-}}$

★ ʟᴇᴛ ᴛʜᴇ ғɪʀsᴛ ᴛᴇʀᴍ ᴀɴᴅ ᴛʜᴇ ᴄᴏᴍᴍᴏɴ ᴅɪғғᴇʀᴇɴᴄᴇ ᴏғ ᴛʜᴇ ɢɪᴠᴇɴ ᴀ.ᴘ ʙᴇ a ᴀɴᴅ d, ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.

❥︎ Sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ 7 ᴛᴇʀᴍs,

$\boxed{ S _{7} = 49} .$

✰ᴡᴇ ᴋɴᴏᴡ,

$\boxed{  s = \frac{n}{2} [ 2a +( n-1 )d ] }$

$→  \frac{7}{2} \: (2a + 6d \: = 49$

$→  \frac{7}{2} \: \times 2(a + 3d) = 49$

$\boxed{→ a + \: 3d = 7 } ........(i)$

❥︎ sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ 17 ᴛᴇʀᴍs ,

$\boxed{ S_{17} \: = 289}$

$→ \frac{17}{2} \: (2a + 16d) = 289$

$→ \frac{17}{2} \times 2(a + 8d) = 289$

$→ \: a + 8d = \: \frac{289}{17} = 17$

$→ \boxed{ a + 8d = 17 } .........(ii)$

★ sᴜʙᴛʀᴀᴄᴛɪɴɢ (ii) ғʀᴏᴍ (i) , ᴡᴇ ɢᴇᴛ

5d =10

$→   \boxed{ d = 2 }$

★ sᴜʙsᴛɪᴛᴜᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ d ɪɴ (i) ᴡᴇ ɢᴇᴛ

$→  \boxed{ a = 1 }$

✰ɴᴏᴡ ,

sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ n ᴛᴇʀᴍs ɪs ɢɪᴠᴇɴ ʙʏ

$\boxed{  s = \frac{n}{2} [ 2a +( n-1 )d ] }$

$→ \frac{n}{2} [ 2 ×1 +2 ( n-1 ) ]$

$==> \boxed{ n( 1 + n - 1 = {n}^{2} } .$

❥︎ᴛʜᴇʀᴇғᴏʀᴇ , ᴛʜᴇ sᴜᴍ ᴏғ ᴛʜᴇ ғɪʀsᴛ n ᴛᴇʀᴍs ᴏғ ᴛʜᴇ ᴀᴘ ɪs

${n}^{2}$

❥︎ ɦσρε เƭ ɦεℓρร !