Question

The first and last terms of an ap are 1 and 12 respectively find (a) the number of terms in the ap (b) the common difference between them if the sum of its terms is (i) 594(ii) 671 (iii) 847 (iv) 1281?

Answer 1

Step-by-step explanation:

Let a= first term= 1

d= common difference

L= last term= 12

Tn= № of terms

Sn= sum of the nth terms

Tn= a + (n - 1)d

Sn= ½(n)(a + L)≈ ½(n)(2a + (n - 1)d)

(a)(i) When Sn= 594,

½(n)(1 + 12)= 594

½(n)(13)= 594

13n= 594 × 2

13n= 1188

n= 1188/13

n= 91.3846

(ii) When Sn= 671,

½(n)(1 + 12)= 671

½(n)(13)= 671

13n= 671 × 2

13n= 1342

n= 1342/13

n= 103.2308

(iii) When Sn= 847,

½(n)(1 + 12)= 847

13n= 847 × 2

13n= 1694

n= 1694/13

n= 130.3077

(iv) When Sn= 1281,

½(n)(1 + 12)= 1281

13n= 1281 × 2

13n= 2562

n= 2562/13

n= 197.0769

(b) Recall, ½(n)(2a + (n - 1)d)= ½(n)(a + L)

2a + (n - 1)d= a + L

2a + (n - 1)d= a + L

2a - a + (n - 1)d= L

a + (n - 1)d= L •••> Equation 1

(i) When Sn= 594, n= 91.3846;

1 + (91.3846 - 1)d= 12

90.3846d= 12 - 1

d= 11/90.3846

d= 0.1217

(ii) When Sn= 671, n= 103.2308,

1 + (103.2308 - 1)d= 12

102.2308d= 12 - 1

d= 11/102.2308

d= 0.1076

(iii) When Sn= 847, n= 130.3077,

1 + (130.3077 - 1)d= 12

129.3077d= 12 - 1

d= 11/129.3077

d= 0.0851

(iv) When Sn= 1281, n= 197.0769,

1 + (197.0769 - 1)d= 12

196.0769d= 12 - 1

d= 11/196.0769

d= 0.0561

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Answer 2

Step-by-step explanation:

Let a= first term= 1

d= common difference

L= last term= 12

Tn= № of terms

Sn= sum of the nth terms

Tn= a + (n - 1)d

Sn= ½(n)(a + L)≈ ½(n)(2a + (n - 1)d)

(a)(i) When Sn= 594,

½(n)(1 + 12)= 594

½(n)(13)= 594

13n= 594 × 2

13n= 1188

n= 1188/13

n= 91.3846

(ii) When Sn= 671,

½(n)(1 + 12)= 671

½(n)(13)= 671

13n= 671 × 2

13n= 1342

n= 1342/13

n= 103.2308

(iii) When Sn= 847,

½(n)(1 + 12)= 847

13n= 847 × 2

13n= 1694

n= 1694/13

n= 130.3077

(iv) When Sn= 1281,

½(n)(1 + 12)= 1281

13n= 1281 × 2

13n= 2562

n= 2562/13

n= 197.0769

hope it helpful..