मराठी

The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

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प्रश्न

The fourth term of an A.P. is 11 and the eighth term exceeds twice the fourth term by 5. Find the A.P. and the sum of first 50 terms.

बेरीज
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उत्तर

For an A.P.

t4 = 11

`\implies` a + 3d = 11  ...(i)

Also, t8 – 2t4 = 5

`\implies` (a + 7d) – 2 × 11 = 5

`\implies` a + 7d – 22 = 5

`\implies` a + 7d = 27  ...(ii)

Subtracting equation (i) from equation (ii), we get

a + 3d = 11
a + 7d = 27
–   –       –    
– 4d = – 16

d = `(- 16)/(- 4)`

`\implies` d = 4

Substituting d = 4 in equation (i), we get

a + 3d = 11

a + 3 × 4 = 11

`\implies` a + 12 = 11

`\implies` a = –1

∴ Required A.P. = a, a + d, a + 2d, a + 3d, ....

∴ a = –1

∴ a + d = –1 + 4 = 3

∴ a + 2d = –1 + 2(4) = –1 + 8 = 7

∴ a + 3d = –1 + 3(4) = –1 + 12 = 11

Sum of first 50 terms of this A.P.

=`50/2 [2 xx (-1) + 49 xx 4]`

= 25[–2 + 196]

= 25 × 194

= 4850

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पाठ 10: Arithmetic Progression - Exercise 10 (C) [पृष्ठ १४४]

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सेलिना Concise Mathematics [English] Class 10 ICSE
पाठ 10 Arithmetic Progression
Exercise 10 (C) | Q 14. | पृष्ठ १४४
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